CIRCULAR MOTION
Circular Motion 1.
Acceleration of a body moving with constant speed in a circle is : (A) zero
2.
(B) × r
(C)
2 r
(D*) 2 r
A pendulum of length l = 1 m is released from 0 = 60º. The rate of change of speed of the bob at = 30º is (g = 10 m/s2)
(A) 5 3 m/s2
(B*) 5 m/s2
(C) 10 m/s2
(D) 2.5 m/s2
3.
A stone weighing 50 g tied to one end of the string is to be rotated in a horizontal circle of 1 metre with a speed of 5 ms–1. The centripetal force required to do so is _________. [Ans. 1.25 N]
4.
A flywheel makes 600 rpm. The angular speed of any point on the wheel and the linear speed of a point 5 cm from the centre of the wheel are _________. [Ans. 100 cm s–1]
5.
A particle is moving in a circle of radius R in such a way that at any instant the total acceleration makes an angle of 45º with radius. Initial speed of particle is 0. The time take to complete the first revolution is R 2 (A) e 0
R 2 (B*) (1 e ) 0
R (C) 0
2R (D) 0
6.
A particle is revolving in a circle with increasing its speed uniformly . Which of the following is constant? [S-05-06/DPP-61(A-batch)/Q.1] (A) centripetal acceleration (B) tangential acceleration (C*) angular acceleration (D) none of these
7.
Two particles A and B revolve concentrically in a horizontal plane in the same direction. The time required to complete one revolution for particle A is 3 min. while for particle B is 1 min. The time required for A to make one revolution relative to B is : [S-05-06/DPP-61(A-batch)/Q.3] (A) 3 min (B) 1 min (C*) 1.5 min (D) none
8.
The square of the angular velocity of a certain wheel increases linearly with the angular displacement during 100 rev of the wheel's motion as shown . Compute the time t required for the increase . [S-05-06/DPP-61(A-batch)/Q.6]
[Ans.
40 ] 7
CIRCULAR MOTION
9.
A ring of radius 1 m rotates about z axis as shown in figure . The plane of rotation is xy . At a certain instant the acceleration of a particle P (shown in figure) on the ring is ,
a 3 i 4 j m/s2 . At that instant angular acceleration of the ring is _______ & the angular velocity is _______ .
[S-05-06/DPP-61(A-batch)/Q.7]
[ Ans : 3 rad/s2 , 2 rad/s2 ] 10.
A particle starts moving at t = 0 in a circle of radius R = 2 m with constant angular acceleration of = 3 rad/sec2. Initial angular speed of the particle is 1 rad/sec . At the instant when the angle between the acceleration vector and the velocity vector of the particle is 37º, calculate ; [S-05-06/DPP-61(A-batch)/Q.8] (a) the value of ‘t’ at this moment (b) magnitude of the acceleration of the particle (c) distance travelled by the particle upto this moment. 1 5 sec (b) 7.5 (c) [Ans. (a) ] 6 12
11.
A particle moving on the inside of a smooth sphere of radius r describing a horizontal circle at a distance r/ 2 below the centre of the sphere. What is its speed? [S-05-06/DPP-61(A-batch)/Q.9] [ Ans:
3 gr ] 2
COMPREHENSION : (12 - 17) Test your comprehension skills. Below given is a passage which you have to read and answer the questions given after the examples. For this you can refer to circular motion chapter of your books. Stick to the time limit given for the test. Angular Variables Suppose a particle P is moving in a circle of radius r (figure). Let O be the centre of the circle. Let O be the origin and OX the X-axis. The position of the particle P at a given instant may be described by the angle between OP and OX. We call the angular position of the particle. As the particle moves on the circle, its angular position changes. Suppose the particle goes to a nearby point P' in time t so that increases + . The angular displacement thus is . The rate of change of angular position is called angular velocity. Thus, the angular velocity (denoted by ) We can say that angular velocity =
Angular displaceme nt Time taken
P
= Lim
t 0
P
O
X
.
d t dt
CIRCULAR MOTION
The rate of change of angular velocity is called angular acceleration. Thus, the angular acceleration is =
Lim t0
d d2 2 . = t dt dt
Same as that we have done for linear motions where x is the position, V = a=
dx is the velocity and dt
dv d is the acceleration; In circular motion we have as the angular position, = , the angular dt dt
velocity, and =
d as the angular acceleration. dt
Linea r Quantities
An gular Qu antities
Displacem ent
Displacement
x
Angular dis placement
Linear Ve locity
Linear Displacement Time Taken
V
Angular Ve locity
=
Linear Acceleration
Change in Linear Velocity Time taken
a
Angular Acceleration
=
Ang ular displacem ent
d dt
Angular Displaceme nt Time taken
d dt
Change in Angular Velocity Time taken
Curiously all the three formulae applied in linear kinematics can be applied in circular kinematics. V = u + at f = i + t 1 2 at 2 V2 – U2 = 2a (S)
S = ut +
1 2 t 2 f2i 2 = ()
= i t +
The above formulae are valid for constant acceleration and constant angular acceleration only. The relation between the circular quantities () and linear quantities (x, v, a) can be derived. We know that length of arc = radius × angle eg : length of circumference = r × (2 ) for any angle length of arc is x=r× dx d Similarly V = =r = r dt dt
x
x = r
V = r
r
dv d a =r = r = dt dt r Calculate the angular velocity of the hour, minute and second hand of a wall clock. Also calculate the angular displacement of the minute hand in 20 minutes. Calculate the angular acceleration of the second hand.
Similarly a = Eg.
Ans. Angular displacement of the minute hand is
2 × 20 radian. 60
2 2 Angular velocity of hour hand = 12 hrs. 12 60 60 rad sec–1
Angular velocity of minute hand =
2 2 rad sec–1 60 60 3600
Angular velocity of second hand =
2 rad sec–1 60
d dt Here the angular velocities are constant and therefore there is no angular acceleration
Angular acceleration =
d =0 dt A stone is tied to a 1.5 m string and whirled in a horizontal circle such that its angular acceleration is 2 rad/
hence = Eg.
CIRCULAR MOTION 2
Sol.
sec . Its initial angular velocity was 2 rad/sec. Find how many revolutions will it make in 10 seconds. Also find the final angular velocity. The can be solved by the equation of motion i.e. 1 2 t 2 initial angular velocity. angular acceleration. tThe time for which it is whirled.
= t + where
Here i = 2 rad sec–1 , = 2 rad sec–2 and t = 10 sec.
= 2 × 10 +
1 2 × 102 = 120 radians. 2
We know that one revolution is 2 radians 120 60 revolutions = = 19 revolutions (approximately.) .) 2 also we know that f = i + t (where f is final angular velocity) f = 2 + 2 × 10 = 22 rad/sec.
120 radians means that
Now Answer the following questions based on the information and examples given above. Time Limit : 25 Min. 12. What is angular displacement ? what are its units ? What is the angular displacement of a particle moving in a circle in : [4 Marks] (i) One rotation (ii) Half rotation (iii) Quarter rotation 13. A car goes around a traffic circle in 60 seconds. What is the angular displacement in 10 seconds ? (Give your answer in radians) ? What is the angular velocity in rad/sec. [4 Marks] 14. Find the angular velocity of the earth around the son. (Assume it to have a circular path and a non-leap year). Similarly find the angular velocity of the moon (Moon takes 29 days to complete one revolution of earth). Give your answer in rad/sec. [4 Marks]
15.
2 2 , m = 365 24 60 60 29 24 60 60 A fan rotating at an angular velocity of 20 radian/sec. is switched of f. It is observed that the fan stops in 20 seconds. Find the angular deceleration of the fan and the number of revolutions made by it till it stops. [4 Marks] Ans. = rad/sec2. 10 revolutions
16.
If a body moving in a circle of radius 2 m has a velocity of 4 m/s. Find its angular velocity.
Ans. e =
Ans. =
V 4 = = 2 rad/sec. r 2
[4 Marks]
17.
Find the acceleration of a particle placed on the surface of the earth at equator due to earth’s rotation. The diameter of earth = 12800 km. the period of earth’s rotation = 24 hrs. [4 Marks]
18.
Two bodies A & B separated by a distance 2 R are moving counter clockwise along a circular path of radius R each with uniform speed v . At time t = 0 , A is given a constant tangential acceleration a= (i) (iii)
72 v 2 . Find : 25 R the time lapse for the two bodies to collide angular velocity of A
[Ans : (i) 19.
[S-05-06/XI(A-batch)DPP-61/Q.10]
5R sec 6v
(ii)
11 6
(iii)
17 v 5R
(ii) (iv) (iv)
the angle covered by A radial acceleration of A .
289 v 2 ] 25 R
A mass m is suspended from the fixed point P by a light inextensible string of length l and describes a horizontal circle under the action of no forces other than its weight and tension in the string. The tension in the string is: [S-05-06/XI(A-batch)DPP-62/Q.1] (A*) proportional to the square of angular velocity with which the particle describes the horizontal circle (B) proportional to the square root of angular velocity with which the particle describes the horizontal circle (C) proportional to the angular velocity with which it describes the circle (D) independent of the angular velocity with which it describes the circle.
CIRCULAR MOTION
20.
Amplitude of simple pendulum is 60º. Find the tension in string when string makes an angle of 30º with vertical. [S-05-06/XI(A-batch)DPP-62/Q.2] (A) mg
21.
(B) 3 3 2 mg
3 3 1 2
(C*) mg
(D) none of these
A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it, reaches a position, where the string is horizontal, is [S-05-06/XI(A-batch)DPP-62/Q.3] (A)
22.
3 2
u2 2gL
(B)
2gL
(C)
u2 gL
(D*)
2(u 2 gL )
A small block slides with velocity 0.5 gr on the horizontal frictionless surface as shown in the Figure. The block leaves the surface at point C. The angle in the Figure is [S-05-06/XI(A-batch)DPP-62/Q.4]
(A) cos1 (4/9) 23.
(B*) cos1(3/4)
(C) cos1 (1/2)
(D) none of these
A heavy particle is hanging from a fixed point by a light inextensible string of length l. It is projected horizontally with speed g . Find the speed of the particle and the angle of string with vertical at the instant of motion when the tension in the string is equal to the weight of the particle. [S-05-06/XI(A-batch)DPP-62/Q.8] [ Ans.: cos = 2/3, V = g / 3 ]
24.
25.
A train rounds an unbanked circular bend of radius 30m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose ? The engine or the rails ? The outer or the inner rail ? Which rail will wear out faster, the outer or the inner rail ? What is the angle of banking required to prevent wearing out of the rails ? [S-05-06/XI(A-batch)DPP-63/Q.6] 3 [ Ans.: 37º 25 tan–1 ] 4 A circular race track of radius 300 m is banked at an angle of 15º. If the coefficient of friction between the wheels of a race car and the road is 0.2, what is the: [S-05-06/XI(A-batch)DPP-63/Q.7] (i) optimum speed of the race car to avoid wear and tear on its tyres (ii) maximum permissible speed to avoid slipping ? [ Ans.: v = 10 30 tan 15 (i) 28.06 m/s (ii) 38.13 m/s ]
26.
Figure on the right shows a rod of length 20 cm pivoted near an end and which is made to rotate in a horizontal plane with a constant angular speed. A ball of mass m is suspended by a string also of length 20 cm from the other end of the rod. If the angle made by the string with the vertical is 30º, find the angular speed of rotation. Take g = 10 m/s2.
[Ans. = 4.4 rad/sec.]
CIRCULAR MOTION
27.
A particle is attached by means of two equal strings to points A and B in the same vertical line and decribe a horizontal circle with a uniform angular speed. If the angular speed of the particle is 2 ( 2g / h) with AB = h, show that the ratio of the tension of the string is 5 : 3.
28.
A point moves in the plane so that its tangential acceleration wt = a & its normal acceleration wn = bt4, where a & b are positive constants & t is time. At the moment t = 0 the point was at rest. Find how the curvature radius R of the point's trajectory & the total acceleration w depend on the distance covered s.
2
4bs2 a3 [ Ans: R = , = a 1 3 ] 2bs a 29.
What is the relation between tangential acceleration aT and the centripetal acceleration aC during non uniform circular motion? (A) aT must be greater than aC (B) aT must be less than aC (C) aT must be equal to aC (D*) None of the above relations is true.
30.
A simple pendulum has a bob of mass m and swings with an angular amplitude . The tension in the thread is T. At a certain time, the string makes an angle with the vertical (). (A) T = mg cos , for all values of (B*) T = mg cos , only for =
1
(C*) T = mg, for = cos 1 (2 cos 1) 3 31.
(D*) T will be larger for smaller values of .
A simple pendulum of length l is set in motion such that the bob, of mass m, moves along a horizontal circular path and the string makes a constant angle with the vertical. The time period of rotation of the bob is t and the tension in the thread is T. (A) t = 2
g
(B*) t = 2
cos g
4 2 m (C*) T = (D) the bob is in equilibrium. t2
32.
A car is taking a turn on a level road. It may be thrown outwards because of the DPP 66_ACJ_05-06 (A) weight (B*) lack of centripetal force (C) reaction of the ground (D) frictional force
33.
A particle moves along a circular path of constant radius. The magnitude of its acceleration is DPP 66_ACJ_05-06
(A) uniform (B) variable (B) zero (D*) such as cannot be predicted from the given information 34.
Which of the following statements about the centripetal and centrifugal forces is correct ?
35.
(A) Centripetal force balances the centrifugal force (B) Both centripetal force and centrifugal force act on the same body (C*) Centripetal force is directed opposite to the centrifugal force (D) Centripetal force is experienced by the observer at the centre of the circular path described by the body A motor cyclist wants to drive on the vertical surface of a wooden 'well' of radius 5m, with a minimum speed
DPP 66_ACJ_05-06
of 5 5 . The minimum value of coefficient of friction between the tyres and the wall of the well must beDPP 68_ACJ_05-06
(A) 0.10
(B) 0.20
(C) 0.30
(D*) 0.40
CIRCULAR MOTION
36.
A motorcyclist of mass m is to negotiate a curve of radius r with a speed v. The minimum value of the coefficient of friction so that negotiatiion may take place safely, isDPP 68_ACJ_05-06 2
(A) v rg
(B*)
v2 gr
(C)
gr v2
(D)
g v 2r
37.
A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1g sticking at the outer end of a blade. How much forcedoes it experience when the fan runs at full speed ? Who exerts this force on the particle ? How much force does the particle exert on the blade along its surface ? [HCV 1/Circular Motion/Exercise/Q.11] [Ans : 14.8N, 14.8 N]
38.
A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure). A smooth pulley of small radius is fastended to the table. Two masses m and 2m placed on the table are conneted through a string over the pulley. Initially the masses are held by a person with the string along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string. [HCV 1/Circular Motion/Exercise/Q.30]
[Ans : 39.
2R 4 2 , m R ] 3 3
A small ring P is threaded on a smooth wire bent in the form of a circle of radius ‘ a ’ and centre O. The wire is rotating with constant angular speed about a vertical diameter XY, while the ring remains at rest relative to the wire at a distance a/2 from XY, then 2 = _______. DPP 63_05-06_5 [ Ans:
2g ] a 3
40.
Three girls Sushma, Rashmi and Priya are on the merrygoround. Sushma and Rashmi occupy diametrically opposite points on a merrygoround of radius r. Priya is on another merrygoround of radius R. The position of the girls at the initial instant are shown in the figure. Consider that the merrygo round touch each other and rotate in the same direction at the same angular velocity , determine the nature of motion of Priya from Rashmi point of view and of Susma from Priya point of view. DPP 63_05-06_8
Ans.
(a) from Rashmi point of view motion of priya (R + r) radius angular velocity = . (b) From priya point of view motion of Sushma (R – r) - radius angular = opposite
41.
To enable a particle describe a circular path what should be the angle between its velocity and acceleration? DPP 66_05-06_18 (A) 0° (B) 45° (C*) 90° (D) 180°
42.
Two particles move on a circular path (one just inside and the other just outside) with angular velocities and 5 starting from the same point. Then : ACJ_DPP 76_05-06
2 when their angular velocities 4
(A)
they cross each other at regular intervals of time
(B*)
are oppositely directed they cross each other at points on the path subtending an angle of 60 oat the centre if their angular velocities are oppositely directed
CIRCULAR MOTION
43.
they cross at intervals of time
(D*)
hey cross each other at points on the path subtending 90 o at the centre if their angular velocities are in the same sense.
A mass M slides down a curved frictionless track in vertical plane, starting from rest. The curve obeys the equation y = x2/2. The tangential acceleration of the mass is: ACJ_DPP 76_05-06 (A) g
44.
if their angular velocities are oppositely directed 3
(C*)
(B)
gx x2 4
(C)
g 2
(D*)
gx x2 1
A heavy particle is projected from a point on the horizontal at an angle 60º with the horizontal with a speed of 10 m/s. Then the radius of the curvature of its path at the instant of crossing the same horizontal is: ACJ_DPP 76_05-06
(A) infinite 45.
(B) 10 m
(C) 11.54 m
(D*) 20 m
Wheel A of radius rA = 10cm is coupled by a belt C to another wheel of radius rB = 25 cm as in the figure. The wheels are free to rotate and the belt does not slip. At time t = 0 wheel A increases it’s angular speed from rest at a uniform rate of /2 rad/sec2 . Find the time in which wheel B attains a speed of 100 rpm.
A-Batch_DPP-60_05-06_9 [ Hint : vA = vB] 46.
[ Ans : 50/3 sec. ]
A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating and is [Q. 3.10_CM] ,dd.ktksfcUnqAij,dfLFkjfpdusrf=kT;kdsv)Zxksysijfp=kkuqlkjfojkekoLFkkesagSAfcUnqBij;gv)Zxksysls laidZNksM+nsrkgSACv)ZxksysdkdsUnzgSrFkkdkslEc)djusokyhlehdj.kgksxh-
[Made 2005, MPS]
Sol.
(A) 3 sin = 2 cos (B) 2 sin = 3 cos (C*) 3 sin = 2 cos (D) 2 sin = 3 cos (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv 2 = mg sin .......... (1) r Applying conservation of energy as the block moves from A to B..
1 mv 2 = mg (r cos – r sin ) 2 Solving 1 and 2 we get 3 sin = 2 cos
47.
A particle of mass m describes a circle of radius r. The centripetal acceleration of the particle is 4/r2. What will be the momentum of the particle ? M.Bank_CM_1.51 2 m nzO;ekudk,dd.krf=kT;kdso`resaxfr'khygSAd.kdkvfHkdsUnzh;Roj.k4/r gSaAd.kdklaosxD;kgksxkA (A) 2
48.
.......... (2)
m r
(B*) 2
m r
(C) 4
m r
(D) none
In the figure shown a lift goes downwards with a constant retardation. An observer in the lift observers a conical pendulum in the lift, revolving in a horizontal circle with time period 2 seconds. The distance between the centre of the circle and the point of suspension is 2.0 m. Find the retardation of the lift in m/s2.
CIRCULAR MOTION
Use
2
= 10 and g = 10
m/s2
[Made 2005 RKV]
M.Bank_CM_2.1
fn,x,fp=kesafy¶Vfu;reanulsuhpsdhvkSjxfrdjjghgSA,dizs{kdtksfy¶VdsvUnjgS]2lSd.MdsvkorZ&dky ls{kSfrto`ÙkesapDdjdkVjgkgSAo`rdsdsUnz,oe~yVdu fcUnqdschpnwjh2.0mgSAfy¶VdkeanuKkrdhft,A mi;ksx esa ys2 = 10 vkSjg = 10 m/s2 [Made 2005 RKV] M.Bank_CM_2.1
Sol.
T = 2
cos geff . = 2
h geff .
geff. = g + a ; put T = 2 a = 10 m/s2. Ans.
Retardation = 10 m/s2
Ans. 10 49.
A semicircular portion of radius ‘r’ is cut from a uniform rectangular plate as shown in figure. The distance of centre of mass 'C' of remaining plate, from point ‘O’ is : GRST_COM_Ex.2_A-4
(A) 50.
2r (3 )
3r (B) 2 ( 4 )
(C)
2r (D*) 3 ( 4 )
2r ( 4 )
A circular plate of diameter d is kept in contact with a square plate of edge d as shown in figure. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be GRST_COM_Ex.2_A-6
//////////////////////////////////////////////////////
d (A) inside the circular plate (C) at the point of contact 51.
d (B*) inside the square plate (D) outside the system
Two particles bearing mass ratio n : 1 are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is: GRST_COM_Ex.2_B-3 2
(A) (n – 1)2 g
52.
53.
n 1 g (B) n 1
2
n 1 g (C*) n 1
n 1 g (D) n 1
Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration a . The centre of mass has an acceleration. GRST_COM_Ex.2_B-6 1 (A) zero (B*) a (C) a (D) 2 a 2 A shell is fired from a canon with a velocity V at an angle with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is GRST_COM_Ex.2_C-11
CIRCULAR MOTION
(A*) 3V cos 54.
(C)
3 V cos 2
(D) V cos
A skater of mass m standing on ice throws a stone of mass M with a velocity of v m/s in a horizontal direction. The distance over which the skater will move back (the coefficient of friction between the skater and the ice is ) : GRST_COM_Ex.2_C-13 (A)
55.
(B) 2V cos
M2 v 2 2 mg
(B)
Mv 2
(C*)
2
2m g
M2 v 2
(D)
2
2m g
M2 v 2 2 m2 2 g
Two blocks of masses m and M are moving with speeds v 1 and v 2 (v 1 > v 2) in the same direction on the frictionless surface respectively, M being ahead of m. An ideal spring of force constant k is attached to the backside of M (as shown). The maximum compression of the spring when the block collides is: GRST_COM_Ex.2_D-1
(A) v1
m k
(B) v2
M k
mM (C*) (v1 – v2) (M m) K
(D) None of above is correct.
56.
A bullet of mass m moving vertically upwards with a velocity 'u' hits the hanging block of mass 'm' and gets embedded in it. The height through which block rises after the collision, assume sufficient space above block : GRST_COM_Ex.2_E-3 (A) u2/2g (B) u2/g (C*) u2/8g (D) u2/4g
57.
A particle of mass m moves with velocity v0 = 20 m/sec towards a wall that is moving with velocity v = 5 m/sec. If the particle collides with the wall elastically. The speed of the particle just after the collision is: GRST_COM_Ex.2_F-11
(A*) 30 m/s 58.
(B) 20 m/s
(C) 25 m/s
(D) 22 m/s
A bullet of mass m strikes a block of mass M connected to a light spring of stiffness k, with a speed V0. If the bullet gets embedded in the block then, the maximum compression in the spring is : GRST_COM_Ex.2_F-16
1/ 2
m 2v 0 2 (A*) (M m) k
59.
1/ 2
Mmv 02 (B*) 2 ( M m ) k
Mv 0 2 (C) 2(M m) k
1/ 2
1/ 2
mv 2 (D) ( M m ) k
A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass. If e is the coefficient of restitution, then ratio of velocity of the first sphere to the velocity of the second sphere after collision will be : GRST_COM_Ex.2_F-17
1 e (A*) 1 e
1 e (B*) 1 e
e 1 (C) e 1
e 1 (D) e 1
60.
If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate of combustion of the fuel will be : GRST_COM_Ex.2_G-2 (A) 10.7 kg/sec (B) 0.07 kg/sec (C) 1.4 kg/sec (D*) 0.7 kg/sec
61.
A particle moves along a circle of radius R with a constant angular speed .Its displacement (only magnitude) in time t will be [M.Bank(07-08)_C.M._1.12] ,dd.kRf=kT;kdso`Ùkesafu;rdks.kh;pkylsxfrdjjgkgSAbldkfoLFkkiutle;esa¼dsoyifjek.k½gksxkA
CIRCULAR MOTION
(A) t
(B) 2 R cos t
(C) 2 R sin t
t 2 [Made PKS 2006, F1-F3]
(D*) 2R sin
Sol.
62.
R 2 R2 x2
t 2 2R A particle is revolving in a circle increasing its speed uniformly. Which of the following is constant?
cos t =
x = 2R sin
2
[M.Bank(07-08)_CM_1.38]
,dd.kdhpkydks,dlekunjlsc<+krsgq,,do`Ùkh;iFkij?kqek;ktkrkgSArksfuEuesalsdkSulhjkf'kfu;rgksxh (A) centripetal acceleration (vfHkdsUnzh; Roj.k) (B) tangential acceleration (Li'kZjs[kh; Roj.k) (C*) angular acceleration (dks.kh; Roj.k) (D)noneof these(buesals dksbZugha) Sol.
63.
d v at a Angular acceleration () = Since, = constant t = r dt magnitude of is constant Also its direction is always constant (perpendicular to the plane of circular motion). whereas, direction of at changes continuously a t is not constant.
A particle is revolving in a circle of radius R with initial speed u. It starts retarding with constant retardation v2 8 R . The number of revolutions it makes in time is : 4 R u
[M.Bank(07-08)_C.M._1.39]
v2 ,dd.kRf=kT;kdso`ÙkesaizkjfEHkdpkyulsxfr'khygSA;gfu;renau4 R dslkFkeafnrgksukizkjEHkdjrkgSA;g
Sol.
8 R le;esa fdruspDdj iwjsdjsxk & v (A) 3 (B) 4 Initial Velocity = V = u (say) Velocity at time t = v
dv v2 =– dt 4R
–
dv
v
2
=
dt
4R
1 =0+C u
1 1 1 = t+ 4R u v
= dt
ut 4R 4 Ru ds
ut 4R = 4Ru 1 1 . loge (ut + 4R) = s+C u 4Ru 1 . loge ( 0 + 4R) = 0 + C u
(C*) 2
(D)noneof thesemijksDresalsdksbZugha
CIRCULAR MOTION
ut 4R S = loge 4 R 4R For
t=
8R u
8R 4R u S = loge u 4R 4R 4 R
64.
S = (4R) loge3 = 4R (about)
Number of revolution = 2
A particle of mass m describes a circle of radius r. The centripetal acceleration of the particle is 4/r2. What will be the magnitude of momentum of the particle? [M.Bank_CM._1.43] mnzO;ekudk,dd.krf=kT;kdso`resaxfr'khygSAd.kdkvfHkdsUnzh;Roj.k4/r2gSaAd.kdklaosxdkifjek.kD;k
gksxkA m r
(A) 2
Sol.
v2 =
(B*) 2
r
m
(C) 4
r
p=
4 m2 r
2m r
Two bodies having masses 10 kg and 5 kg are moving in concentric orbits of radii 4 and 8 such that their time periods are the same. Then the ratio of their centripetal accelerations is nksoLrq,saftudsnzO;eku10kgrFkk5kggS]os4rFkk8f=kT;kdsladsUnzh;o`Ùkh;d{kesablrjgpDdjdkVjgsgSrkfd mudk vkorZ&dky ,d tSlk gSA muds vfHkdsUnzh; Roj.kksa dk vuqikr gS & [M.Bank(07-08)_CM_1.11] (A*)
1 2
(B) 2
1 8 [JPNP IIT-Phy./Page-356/Q.37]
(C) 8
(D)
[Q.8/DPP-39/F1-F3]
Sol.
2r1 2r2 v1 r1 1 = = = v1 v2 v2 r2 2
v12 / r1 v 22 / r2
66.
(D) none
4 r
m2 v 2 =
65.
m
2
r2 v1 1 1 = . r = . 2 = 4 2 1 v2
A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration
of a particle P (shown in figure) on the ring is (6 i 8 j ) m/s2. At that instant angular acceleration of the ring is _______ & the angular velocity is _______. Radius of the ring is 2m. fp=kkuqlkj ,doy; Zv{kds lkis{k?kw.kZu djrhgSA ?kw.kZuxfr dkryxygSAfdlh {k.koy;ijfLFkr ,dd.k P
( fp=kkuqlkj)dkRoj.k (6 i 8 j )m/s2 gSA bl{k.koy;dk dks.kh;Roj.k _________gSrFkk dks.kh; osx________
gSA¼oy;dhf=kT;k2ehgSa½
[M.Bank(07-08)_CM._1.16]
CIRCULAR MOTION
Sol.
67.
2 Ans. : –3rad/sec kˆ , – 2 rad/sec kˆ . 2 (2) = 8 = 2 2=6 =3
A car initially traveling eastwards turns north by traveling in a quarter circular path of radius R metres at uniform speed as shown in figure. The car completes the turn in T second. [M.Bank_C.M._1.3] (a) What is the acceleration of the car when it is at B located at an angle of 37. Express your answers in terms of unit vectors ˆi and ˆj (b) The magnitude of car's average acceleration during T second period. [Made MPS - 2005]
,ddkjizkjEHkesaiwoZfn'kkesaxfrdjrsgq,,d&pkSFkkbZo`Ùkh;iFkesaxfrdjrsgq,lsmÙkjfn'kkesaeqM+tkrhgSAo`Ùkh; iFkdhf=kT;kRrFkkiFkesapky,dlekugSAdkjiFkdksTlsd.MesaikjdjrhgSA [M.Bank(0708)_C.M._1.3] (a) dkj dkRoj.kD;kgksxktc;gAls37dks.kijfLFkrfcUnqBijgksA[mÙkjlfn'k ˆi rFkk ˆj esanhft,A ] (b)T lsd.MesadkjdkvkSlrRoj.kdkifjek.kD;kgksxk [Made MPS - 2005]
Sol.
Speed of car is v =
R m/s 2T
.....
2 R v2 (a) The acceleration of car is = at B and is directed from B to O. 4 T2 R Acceleration vector of car at B is 2 R v2 ˆ ˆ = ( – sin 37° + cos 37° ) = (– 3 ˆi + 4 ˆj ) m/s2 j i a 20T2 R (b) The magnitude of average acceleration of car is in time T is R v C VB 2v = = m/s2 2 T2 T T
68.
A car mov es around a curv e at a constant speed. W hen the car goes around the arc subtending 60° at the centre, then the ratio of magnitude of instantaneous acceleration to average acceleration over the 60° arc is : ,d dkj oØ ij fLFkj pky ls xfreku gSA tc dkj dsUnz ds ifjr 60°ds dks.k dk pki r; djrh gS] rc 60° ds dks.k
CIRCULAR MOTION
dspkidsfy,rkR{kf.kd,oavkSlrRoj.kksadkvuqikrgS&
[M.Bank(07-08)_C.M._3.17]
[Made 2006, JKS, GRSTU]
(A*)
Sol.
69.
3
(B)
6
(C)
2 3
(D)
5 3
| V | =
v 2 v 2 2v 2 cos 60 =v 3 v2 | v | v aav = = = R t t
v2 ai = R
;
ai v 2 R = a av R 3 v2
=
3
The velocity and acceleration vectors of a particle undergoing circular motion are v = 2ˆi m/s and a = 2ˆi + 4ˆj m/s2 respectively at an instant of time. The radius of the circle is [Made VSS, 2006-GRST] o`Ùkh; xfr dj jgs d.k dk fdlh le; ij osx vkSj Roj.k Øe'k% v = 2ˆi m/svkSj a = 2ˆi + 4ˆj m/s2 gSA o`Ùkh; iFk dh
f=kT;kgksxh– Sol.
[M.Bank(07-08)_C.M._1.10] (A*) 1m (B) 2m (C) 3m (D) 4m It can be observed that component of acceleration perpendicular to velocity is
osxdsyEcor~Roj.kds?kVdgS& ac = 4 m/s2
70.
radius =
v2 (2)2 = = 1 metre. ac 4
STATEMENT-1 : If a body is thrown vertically upwards (from the earth surface) the distance covered by it in the last second of upward motion is about 5 m irrespective of its initial speed. STATEMENT-2 : The distance covered in the last second of upward motion is equal to that covered in the first second of downward motion when the particle is dropped. oDrO;-1: ;fn fdlh,d oLrqdks lh/kk Å/okZ/kjÅij dh vksjQsadk tkrkgS(i`Fohdhlrg ls) rks Åijtkus dhxfr
dsvfUre,dlsd.Mesar;dhxbZnwjhyxHkx5mgksrhgSpkgsizkjfEHkdosxdqNHkhgksA oDrO;-2 : Åijtkus dhxfr dsvfUre 1lsd.M esar; dhxbZ nwjh]fdlh d.kdks NksM+usij mlds}kjk uhpsdh xfr dsizFkelsd.Mesar;dhxbZnwjhdscjkcjgksrhgSA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (A*) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ; (D) oDrO;-1 vlR; gS] oDrO;-2 lR; gS Sol.
(M oderate)
71.
A body moves along an uneven surface with constant speed at all points. The normal reaction of the road on the body is:
,d oLrq ,d vleku lrg ij lHkh fcUnvksa ij fu;r pky ls py jgh gSA lM+d }kjk oLrq ij vfHkyEc çfrfØ;k gS &
CIRCULAR MOTION
[M.Bank(07-08)_C.M._7.10]
(A*) maximum at A (A*)A ij vf/kdre 72.
(B) maximum at B (B) B ij vf/kdre
(C) minimum at C (C) C ij U;wure
(D) the same at A, B & C (D) A, B o C ij leku
An aeroplane flying at constant speed 115 m/s towards east, makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to complete. The magnitude of the centripetal acceleration during the turn, is [Made A.K.S. sir] [M.Bank(07-08)_C.M._2.30] ,dok;q;ku115m/sdhpkylsiwoZfn'kkesamM+jgkgSA;g/khjslso`Ùkh;iFkesaxfrdjrsgq,nf{k.kfn'kkesaeqM+tkrk gSA;gbleksM+esa15lsd.MysrkgSrkseksM+dsnkSjkuvfHkdsUnzh;Roj.kD;kgksxkA (A)
23 m/s2 8
(B)
46 m/s2 3
(C*)
23 m/s2 6
(D) none of thesebuesa ls dksbZ
ugha Sol.
N
The turn is 1/4 of a circle. Therefore T = 60s T=
TV 2 r 2 V
E
23 V 2 2V = m/s2 S 6 r T A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle is rotated about diameter AB as axis with a speed as shown in figure. The bead P is at rest w.r.t. the circular ring in the position shown. Then 2 is equal to: ,dfpdusrkjdksaf=kT;kdsÅ/okZ/kjo`ÙkesaeksM+rsgSA,deudkP,rkjijfQlyldrkgSAo`Ùkfp=kkuqlkjO;klAB dslkis{kpkyls?kw.kZudjrkgSAfp=kesfn[kkbZfLFkfresaeudkPo`Ùkkdkjoy;dslkis{kfLFkjgSArks2cjkcjgS& [M.Bank(07-08)_C.M._2.15]
and 73.
(A)
Sol.
2g a
(B*) a 2a = 60º N cos60º = mg
2g a 3
(C)
g 3 a
(D)
As ; cos =
2 a N sin60º = m 2
tan60º =
2 =
2g 2 a
2g a 3
w A a O
N
B
mg
2a g 3
CIRCULAR MOTION
74.
75.
In the motorcycle stunt called "the well of death" the track is a vertical cylindrical surface of 18 m radius. Take the motorcycle to be a point mass and = 0.8. The minimum angular speed of the motorcycle to prevent him from sliding down should be: [M.Bank(07-08)_C.M._2.14] ,dekSrdsdq,esaiFk]m/okZ/kj18eh-f=kT;kdhcsyukdkjlrggSAeksVjlkbZfdylokjdksfcUnqnzO;ekuekusrFkk = 0.8gSArks eksVjlkbfdy lokjdh U;wuredks.kh; pkyD;k gksuhpkfg, rkfdog uhpsu fQlysA (A) 6/5 rad/s (B*) 5/6 rad/s (C) 25/3 rad/s (D) none of these The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is fp=kesa,dfcUnqnzO;ekudhxfrdsçkjfEHkd{k.kdsRoj.krFkkosxn'kkZ;sx;sgSA;fnoLrqdk(fcUnqnzO;eku) Roj.klfn'kfu;rjgrkgSrksoLrqdsiFkdhU;wureoØrkf=kT;kgS& [MB_Q. 7.3] BM_C.M._158
[M.Bank(07-08)_C.M._7.3] (A) 2 meter Sol.
[Made 2005, MPS]
(B) 4 meter
(C*) 8 meter (D) 16 meter. [Q.158/RK_BM/Circular Motion] The acceleration vector shall change the component of velocity u|| along the acceleration vector. Roj.klfn'kdslekUrjosxdk?kVdu|| Roj.klfn'k}kjkifjofrZrgksrkgSA
v2 r= an Radius of curvature rmin means v is minimum and an is maximum. This is at point P when component of velocity parallel to acceleration vector becomes zero, that is u|| = 0. oØrkf=kT;k rmin vFkkZr~v U;wurerFkkan vf/kdreAPfcUnqijRoj.klfn'kdslekUrjosxdk?kVd'kwU;gSaAvFkkZr~ u|| = 0
u2 42 = = 8 meters. a 2 An Object follows a curved path. The following quantities may remain constant during the motion : [Bank Obj_CM_80] [M.Bank(07-08)_CM_1.32] ;fnfudk;oØiFkijxfrdjjghgksrksfuEuesalsdkSulhjkf'kxfrdsnkSjkufu;rjgsxh: (A*) speed (B) velocity (C*) acceleration (D*) magnitude of acceleration (A*) pky (B)osx (C*)Roj.k (D*)Roj.kdkifjek.k
76.
77.
R=
A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length . The system is rotated about the other end of the spring with an angular velocity in gravity free space. The increase in length of the spring is : [M.Bank(07-08)_C.M._5.5] mnzO;ekuds,dd.kdksvladqfpryEckbZokyhgYdhfLçaxds,dfljslstksM+ktkrkgSAvcfudk;dksfLçaxdsnwljs
fljsdslkFkdks.kh;osxlsux.;xq:Ro{ks=kesa?kqek;ktkrkgSAfLçaxesafoLrkjgksxk&
CIRCULAR MOTION
(A)
m 2 k
(B*)
kx
Sol.
m 2 k m 2
(C)
m 2 k m 2
(D)noneof these buesalsdksbZugha
2 m(+ x) (pseudo force)
m
kx = m2 + m2 x (k – m2) x = m2 x=
m 2 k – m 2
Ans. (B) 78.
A weightless rod of length 2 carries two equal masses 'm', one secured at lower end A and the other at the middle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C. What horizontal velocity must be imparted to the mass at A so that it just completes the vertical circle. 2yEckbZdhnzO;ekujfgrNM+lsnks'm'nzO;ekudsd.kNM+dsfuEureArFkke/;fcUnqBijtqM+sgq,gSANM+Å/okZ/ kjryesafLFkrfcUnqClsxqtjusokyh{kSfrtv{klsxqtjrhgS?kqek;ktkrkgSAAfcUnqijfLFkrnzO;ekudksfdrus{kSfrt osx fn;k tk,rkfd ;gBhd Å/okZ/kjo`Ùk r; dj ldsA [Made MPS - 2005] [ M . Ban k( 0 708)_C.M._3.5]
Sol.
Let the initial velocity given to the mass at A be u. Then the velocity of mass at B is u/2 As the system moves from initial the final position Increase in potential energy is = 4 mg + 2mg 2
1 1 u 5 2 Decrease in kinetic energy = mu m = mu2 2 2 2 8
From conservation of energy 5 mu2 = 6 mgl 8
79.
or
u=
48 g 5
A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s2 in a circular path. If it slips when its total acceleration becomes 1 m/s2, then the angle through which it would have turned before it starts to slip is : [Made CSS 2006, GRST] [M.Bank(07-08)_C.M._2.32] 2 ,dd.k,do`ÙkkdkjiFkesa0.6m/sec dsfu;rLi'kZjs[kh;Roj.klsxfrdjukizkjEHkdjrkgSAtcbldkdqyRoj.k 1m /sec2 gkstkrkgSarks;gfQlytkrkgSAfQlyukizkjEHkdjuslsBhdigysrdblds}kjk?kwekgqvkdks.kgksxk
CIRCULAR MOTION
& (A) 1/3 rad Sol.
Sol.
(D) 2 rad
2 = 2 + 2 = 0 2 = 2 2R = 2 (R) ac = 2R = 2at 1=
(C) 4/3 rad
a 2t a 2c
aNet =
so
80.
(B*) 2/3 rad
0.36 (1.2 )2
1 – 0.36 = (1.2 )2
0.8 1.2
2 radian 3
A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle (string is horizontal), with a uniform angular speed of 2 rad/sec. Then the tension in the string will be : 1kgnzO;ekuds,diRFkjdks1myEckbZdhjLlhds,dfljslsck¡/kktkrkgStks{kSfrto`Ùkesa,dlekudks.kh;pky 2rad/sec.ls?kqejghgS(jLlh{kSfrtgS) rksjLlhesarukogksxk&KCET_1998 [M.Bank(07-08)_C.M._2.33] (A*) 4 N (B) 8 N (C) 1 N (D) 2 N Given : Mass of the stone m = 1 kg Radius of the circle r = 1 m Angular velocity 2 rad / sec The relation for the tension is given by m 2 r 1 22 1
81.
4N
A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are a = 2 ˆi + 3 ˆj m/s2 and P = 6 ˆi – 4 ˆj kgm/s. Then the motion of the particle is ,d d.k o`Ùkkdkj iFk ij xfr dj jgk gSA fdlh {k.k Roj.k rFkk laosx lfn'k Øe'k%a = 2 ˆi + 3 ˆj m/s2 rFkkP = 6 ˆi – 4 ˆj kgm/sgS rks d.k dh xfr gS & (A) uniform circular motion (C) circular motion with tangential retardation (A) ,d leku o`Ùkh; xfr (C)Li'kZjs[kh;eanudslkFko`Ùkkdkjxfr
Sol.
[M.Bank(07-08)_C.M._1.6] (B) circular motion with tangential acceleration (D*) we cannot say anything from a and P given here. (B)o`ÙkkdkjxfrLi'kZj s[kh;Roj.kdslkFk (D*) fn;s x;sa rFkkP ls dqN ugha dgk tk ldrkA [MB_Q. 1.6] [Made 2005, MPS] The nature of the motion can be determined only if we know velocity and acceleration as function of time. Here acceleration at an instant is given and not known at other times so D.
gy
osxrFkkRoj.kdkdsoyle;dkQyu gksusdhfLFkfresaghxfr dhçd`frKkrdhtkldrhgSA;gkafdlh{k.kRoj.k fn;kgS rFkkvU; le;Kkr ughagS blfy,mÙkjDgksxkA
82.
A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60o with the vertical. Then ,dHkkjhd.k1.6myEckbZdksjLlhds,dfljsijtksM+ktkrkgSrFkknwljkfljktM+Ro¼fLFkj½gSA;g,dnksyd(conical pendulum)dh rjg xfr djrk gSA ftldk Å/okZ/kj ls dks.k60o gSA [M.Bank(07-08)_C.M._1.50] (A*) its period of revolution is
bldk vkorZdky
4 sec. 7
4 sec. lSd.M gksxkA 7
(B*) the tension in the string is doubled the weight of the particle
jLlhesarukod.kdsHkkjdknqxqukgksxkA
CIRCULAR MOTION
(C*) the velocity of the particle = 2.8
3 m/s
d.k dk osx = 2.8 3 m/sgksxkA (D*) the centripetal acceleration of the particle is 9.83 m/s2. d.kdkvfHkdsUnzh;Roj.k9.83m/s2 gksxkA T cos 60o
o
60 o
60
/ 2
T 3/2
V
Sol.
o
T sin 60
mg
2
mv T 3 = ( 3 / 2 ) 2 T = mg 2 Hence T = 2 mg , So (B) holds From (1) & (2) V2 = 3 g/2
........(1)
.......(2)
3 9 .8 1 .6 2
V=
V = 2.8
3 m/s2 . So (C) hold
( 3 g / 2)
ac = V2/r = t=
( 3 / 2)
=
3
(D) holds. 2 3 /2 2 r = ( 3 g / 2) v
t = 4/7 (A) holds. 83.
A particle A moves along a circle of radius R = 50 cm so that its radius vector r relative to the point O (Fig.) rotates with the constant angular velocity = 0.40 rad/s. Find the modulus of the velocity of the particle and the modulus and direction of its total acceleration. ,dd.kR=50cmf=kT;kdso`Ùkesa?kqejgkgSAftldkf=kT;h;lfn'kOdslkis{kr gSAtks¼f=kT;h;lfn'k½fu;rdks.kh; osx =0.40rad/sls?kqerkgSAosx dkifjek.krFkkdqyRoj.kdkifjek.kofn'kkKkrdjksA [Irodov_1.43]
[Ans : v = 2R = 0.40 m/s, w = 4R2 = 0.32 m/s2]
CIRCULAR MOTION
R
w
R
Sol.
V=R
V
C
R
d (20) = 2R = 2 (0.4) (.5) dt
V = 0.4 m/s at = o (Since v is not changing) Hence, a = aC = 84.
(0 . 4 ) 2 V2 = = 0.32 m/s2 0.5 R
The square of the angular velocity of a certain wheel increases linearly with the angular displacement during 100 rev of the wheel's motion as shown. Compute the time t required for the increase. ,difg;sdhxfrdsnkSjkumldsdks.kh;osx dkoxZ]dks.kh;foLFkkiudslkFkjs[kh;:ilsifg;sds 100pDdj ds nkSjku fp=kkuqlkj cnyrk gSA bldh bl o`f} ds fy, yxs vko';d le; t dh x.kuk dhft, & [M.Bank(07-08)_CM_1.22]
40 sec. ] 7 A 40 kg mass, hanging at the end of a rope of length , oscillates in a vertical plane with an angular amplitude of 0. What is the tension in the rope, when it makes an angle with the vertical ? If the breaking strength of the rope is 80 kg f, what is the maximum angular amplitude with which the mass can oscillate without the rope breaking ? ,d 40kgnzO;ekudh ckWy tks yEckbZdh jLlhls Å/okZ/kjry esa0 dks.kh;vk;ke lsnksyu xfrdjrh gSA tc ;g Å/okZ/kjlsdks.kcukrhgSrksjLlhesarukoD;kgksxkA;fnjLlhdkvf/kdreruko80kgcygksrks,vf/kdredks.kh; foLFkkiuD;kgksxkftlls ckWyfcukjLlhVwVsnksyuxfrdjlds? Pg173_6 CM_IIT 1978
[ Ans : 85.
[M.Bank(07-08)_CM_3.70]
Sol.
(b)
The situation is shown in figure. (a) From figure h = (cos – cos 0) and 2 = 2gh = 2g (cos – cos 0) ....... (1) Again T – mg cos = m2 / ....... (2) Substitting the value of 2 from eq. (1) in eq. (2) we get T – mg cos = m {2g (cos – cos 0) /} or T = mg cos + 2mg (cos – cos 0) or T = mg (3 cos – 2 cos 0) or T = 40g (3 cos – 2 cos 0) newton Ans. T = 40 (3 cos – 2 cos 0) kg f.
T h
mg
Let 0 be the maximum amplitude. The maximum tension T will be at mean position where = 0. Tmax = 40 (3 – 2 cos 0) But Tmax = 80
CIRCULAR MOTION
Solving we get 0 = 60° 86.
Ans. 0 = 60°
Two particles tied to different strings are whirled in a horizontal circle as shown in figure. The ratio of lengths of the strings so that they complete their circular path with equal time period is:
jLlhlscU/ksnksd.kksadksfp=kkuqlkjo`ÙkkdkjiFkesa?kqek;ktkrkgSAjfLl;ksadhyEckbZdkvuqikrgSftllsosviuko`Ùkh; iFkleku le;vUrjky esaiwjk djrsgS & [M.Bank(07-08)_CM._1.52]
[Old RRB Q. 86]
(A)
3 2
(B*)
2 3
(C) 1
(D) None of these [Made 2006, SNS, GRSTX]
Sol.
87.
since pwafd
T = 2
L cos g
T1 = T2 L1 cos1 = L2 cos2
L1 cos 2 L 2 cos 1
=
cos 45 cos 30
L1 2 L2 3 If the apparent weight of the bodies at the equator is to be zero, then the earth should rotate with angular velocity M.Bank_C.M./Graviation_4.1/4.5 [M.Bank(07-08)_C.M._4.1]
vxjfo"kqorjs[kkijoLrqvksadkvkHkklhHkkj'kwU;gksrksi`Fohdksfdldks.kh;osxls?kweukgksxk& (A*)
g rad/sec R
(B)
2g rad/sec R
(C)
g rad/sec 2R
(D)
3g rad/sec 2R
g R
Sol.
mg = m2 R , =
88.
Two particles move on a circular path (one just inside and the other just outside) with angular velocities and 5 starting from the same point. Then nksd.k,do`Ùkesa(,dBhdvUnjrFkknwljkBhdckgj)rFkk5dks.kh;osxls,dghfcUnqls'kq:djrsgq,o`Ùkesa
xfrdjjgsgSrks
[M.Bank(07-08)_C.M._1.35]
(A) they cross each other at regular intervals of time
nksauksT=
2 when their angular velocities are oppositely directed 4
2 le;kUrjkyijfeysaxs;fnbudkdks.kh;osxfoijhrfn'kkesagksA 4
(B*) they cross each other at points on the path subtending an angle of 60oat the centre if their angular velocities are oppositely directed nksauksiFkijfLFkrfcUnqtksdsUnzls60o dkdks.kcukrkgSijfeysaxs;fnbudkdks.kh;osxfoijhrfn'kkesagksA (C*) they cross at intervals of time 3 if their angular velocities are oppositely directed
nksauks T=
le;kUrjkyijfeysaxstcbudkdks.kh;osxfoijhrfn'kkesagksA 3
CIRCULAR MOTION o
(D*) they cross each other at points on the path subtending 90 at the centre if their angular velocities are in the same sense. nksauksiFkijfLFkrfcUnqtksdsUnzls90odkdks.kcukrkgSijfeysaxs;fnbudkdks.kh;osx,dghfn'kkesagksA 89.
A circular curve of a highway is designed for traffic moving at 72 km/h. If the radius of the curved path is 100 m, the correct angle of banking of the road should be given by : ,d o`Ùkh; oØkdkj lM+d dks 72 km/h dh pky ds fy, cuk;k x;k gSA ;fn oØ dh f=kT;kR = 100 m gks rks lM+d dk
cadudks.kgksukpkfg,A (A) tan 1 Sol.
(B) tan 1
3 5
(C*) tan 1
2 5
(D) tan 1
1 4
g R tan (20)2 = 10 × 100 × tan
V=
4 2 = =tan–1 (2/5) 10 5 Ans: None
tan =
90.
2 3
[M.Bank(07-08)_C.M._6.3]
A particle is projected horizontally from the top of a tower with a velocity v 0. If v be its velocity at any instant, then the radius of curvature of the path of the particle at the point (where the particle is at that instant) is directly proportional to: [bank new_CM_21] ,dd.kdksehukjls{kSfrtfn'kkesaosxv0lsQSdktkrkgSA;fnfdlh{k.kv osxgks]rksblfcUnqijoØrkf=kT;k(tgka ijml{k.kd.kgS)fuEudslekuqikrhgksxhA [M.Bank(07-08)_C.M._7.5] (A*) v3 (B) v2 (C) v (D) 1/v
Sol. As we know : aC =
v2 (centripetal acceleration). R
From figure ; g sin = 91.
v0 v2 = v R
(Since ; sin i =
v0 ) v
R v3
A wet open umbrella is held upright and whirled about the handle with a uniform angular speed of 21 revolutions in 44 sec. If the rim of the umbrella is a circle of diameter 1m, and height of the rim from the ground is 1.5 m, find the radius of the circle along which the drops of water spun off the rim hit the ground. ,dcjlkrhNkrsdksÅijdhvksjlh/ks[kM+kdjdsbldsgRFksdslkis{kbldks,dlekudks.kh;pky21?kw.kZu44 lSd.M ls?kqek;ktkrkgSA;fnNkrsdhifjf/k,d1mO;kldso`Ùkds:iesagksoifjf/kdh/kjkrylsÅpk¡bZ 1.5mrksikuh dhcwankslscuso`Ùkdhf=kT;kD;kgksxhtksifjf/klsnwjgksusij/kjkryijfxjrhgSA [ M . Ban k( 0 708)_C.M._1.27] [Ans:
Sol.
g.
v2 R
37 m] 40
From projectile motion t=
2h g
CIRCULAR MOTION
V h = 1.5
=V
2h g
=
21x 2 =3 44
r=
1 , 2
V = r =
=
R = radius =
92.
3 2
3 2
3 10 1 2 2
=
9 x3 1 4 x 10 4
=
37 m 40
2
A ring of radius R is placed such that it lies in a vertical plane. The ring is fixed. A bead of mass m is constrained to move along the ring without any friction. One end of the spring is connected with the mass m and other end is rigidly fixed with the topmost point of the ring. Initially the spring is in un-extended position and the bead is at a vertical distance R from the lowermost point of the ring. The bead is now released from rest. ,dRf=kT;kdhoy;dksÅ/okZ?kjryesaj[kkx;kgSAoy;fLFkjvoLFkkesagSAmnzO;ekudk,deudkoy;dhifjf/
kesafcuk?k"kZ.kdsxfrdjldrkgSAfLçaxdk,dfljkeudslsrFkknwljkfljkoy;dslkFkmPprefcUnqtgk¡oy; fLFkjgStksM+ktkrkgSAçkjEHkesafLçaxvfoLFkkfjrgSrFkkeudkoy;dsfuEurefcUnqlsÅ/okZ/kjRnwjhijgSAvceuds dksfojkelsNksM+ktkrkgSA [M.Bank(07-08)_C.M._3.4] (a)
(b)
What should be the value of spring constant K such that the bead is just able to reach bottom of the ring. KfLçaxfu;rkad dkeku D;kgksxk rkfdeudk oy;ds fuEurefcUnq ijigq¡p ldsA The tangential and centripetal accelerations of the bead at initial and bottommost position for the same value of spring constant K. Kdsblhekudsfy,eudsdsLi'kZjs[kh;rFkkf=kT;h;Roj.kçkjfEHkdrFkkfuEurefcUnqfLFkfrdsfy,D;k
gksaxsA
[Made PKS - 2005]
Sol.
(a) Applying conservation of energy between initial and final position is Loss in gravitational P.E. of the bead of mass m = gain in spring P. E.
CIRCULAR MOTION
mg R =
or
K=
1 K (2R – 2 mg
2 2 R)
R (3 2 2 ) (b) At t = 0 at = g ac = 0 at lowest point at = 0 ac = 0 The centripetal acceleration of bead at the initial and final position is zero because its speed at both position is zero. The tangential acceleration of the bead at initial position is g. The tangential acceleration of the bead at lowermost position is zero.
93.
A particle starts from rest at O and moves along a horizontal semi circular track OAB of radius R = 1m as shown in the figure. The rate of change of speed of the particle is constant and equals to 2m/s2. A is a point lying exactly on the middle of semicircular track as shown in figure. When the particle reaches A. Find ,dd.kOlsfojkelsxfrdjrsgq,v/kZo`ÙkkdkjiFkOAB(f=kT;kR =1m)esafp=kkuqlkjxfrdjrkgSAd.kdhpky esaifjorZudhnjfu;ro2m/s2 gSAiFkdse/;esafcUnqAgSAtcd.kAijigq¡prkgSrks [M.Bank(07-08)_Circular Motion_1.4] (a) The magnitude of velocity vector of the particle at the instant.
bl{k.kd.kdsosxlfn'kdkifjek.kgksxkA (b)
Magnitude of acceleration vector of the particle at the instant.
bl{k.kd.kdsRoj.klfn'kdkifjek.kgksxkA (c)
The cosine of angle between acceleration and velocity vector of the particle at that instant bl{k.kRoj.klfn'krFkkosxlfn'kdschpdks.kgksrkscos gksxkA
[Made MPS - 2005]
Sol.
(a) The speed of particle at A is v 2 = u2 + 2as (b)
v=
u = 0, a = 2, s = 2 2
= 2
meters 2
2
v = 2 ˆi m/s The normal acceleration of the particle at A is
v2 = 2m/s2 R tangential acceleration of the particle is = 2m/s2 Total acceleration vector is a = 2ˆi 2 ˆj m/s2 Magnitude of total acceleration is a = 4 4 2 = 2 1 2 m/s2 1 av (c) cos = = a v 1 2 94.
A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating and is fp=kkuqlkj'r'f=kT;kdsfpdusfLFkjv)ZxksysdsfcUnqAls,dd.kfojkekoLFkklsxfrçkjEHkdjrkgSAd.kdkBfcUnq
CIRCULAR MOTION
ij lEidZNwV tkrkgSACv)ZxksysdkdsUnzgSrksrFkk dkstksM+us okyhlehdj.kgS& [MB_Q.3.10] [M.Bank(07-08)_Circular Motion_3.10]
[Made 2005, MPS]
(A) 3 sin = 2 cos Sol.
(B) 2 sin = 3 cos
(C*) 3 sin = 2 cos
(D) 2 sin = 3 cos
(C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv 2 = mg sin .......... (1) r Applying conservation of energy as the block moves from A to B..
gy
1 mv 2 = mg (r cos – r sin ) .......... (2) 2 Solving 1 and 2 we get 3 sin = 2 cos (C) d.k dk lEidZ NwVus ls rqjUr igys d.k dk Bij osx dkVgSA
xksyh; lrg ds yEcor~ d.k ij U;wVu dk fu;e yxkus ij mv 2 = mg sin .......... (1) r CykWddks AlsBrdxfr djusij ÅtkZlaj{k.k dsfu;e ls
1 mv 2 = mg (r cos – r sin ) 2
.......... (2)
1 rFkk 2dks gy djus ij 3 sin = 2 cos 95.
A mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizontal circular path. If is the length of the string from m to the top end of the tube and is angle between this part and vertical part of the string as shown in the figure, then time taken by m to complete one circle is equal to [Made A.K.S. sir]
,dnzO;ekuM gYdhjLlhlsfLFkj:ilsyVdjgkgSrFkkjLlhdknwljkfljkfpduhtM+or~V~;wclsxqtkjusdsckn NksVsnzO;ekumlsca/kkgSAtks{kSfrto`Ùkh;iFkesa?kwejgkgSA;fnmlsV~;wcdsÅijhfljsrdjLlhdhyEckbZvkSj jLlhdsblHkkxrFkkjLlhdsÅ/okZ/kjHkkxdschpdks.k gSrksm }kjk,dpDdjyxkusesao`Ùkiwjkdjusesafy;s x;s le; dk eku gksxk : [M.Bank(07-08)_Circular Motion_2.29]
m M (A) 2 g sin
Sol.
For M to be stationary MfLFkjkoLFkkesagS& T = Mg
(B) 2 g cos
.... (1)
m (C) 2 g M sin
m (D*) 2 g M
CIRCULAR MOTION
Also for mass m, m dsfy, T cos = mg
.... (2) 2
T sin =
mv .... (3) sin
dividing (3) by (2) (3)esa(2)dkHkkxnsusij tan =
v2 v= g sin
g . sin cos
2 sin
Time period vkorZ dky
2 R = = v
g . sin cos
m From (1) and (2) cos = M lehdj.k (1)vkSj (2)ls
then time period vr% vkorZ dky = 2 96.
Tcos
m T M
Tsin
mg
Mg m gM
A rod AB is moving on a fixed circle of radius R with constant velocity ‘v’ as shown in figure. P is the point of 3R from centre of the circle. 5 The velocity of the rod is perpendicular to the rod and the rod is always parallel to the diameter CD. fp=kkuqlkj,dNM+AB,Rf=kT;kdso`Ùkijfu;rosx‘v’lsxfrdjjgkgSAo`ÙkrFkkNM+dkçfrPNsnufcUnqPgSAfdlh 3R {k.ko`ÙkdsdsUnzlsNM+dhnwjhx= gSANM+dkosxNM+dsyEcor~gSrFkkNM+ges'kkO;klCDdslekUrjgSA 5 [M.Bank(07-08)_Circular Motion_1.5]
intersection of the rod and the circle. At an instant the rod is at a distance x =
(a) Find the speed of point of intersection P. çfrPNsnu fcUnqPdh pkyKkr djksA [MB_Q. 1.29] [Made MPS 2005] (b) Find the angular speed of point of intersection P with respect to centre of the circle. çfrPNsnufcUnqPdho`ÙkdsdsUnzdslkis{kdks.kh;pkyKkrdjksA [6]
Sol. (a)
As a rod AB moves, the point ‘P’ will always lie on the circle. its velocity will be along the circle as shown by ‘VP’ in the figure. If the point P has to lie on the rod ‘AB’ also then it should have component in ‘x’ direction as ‘V’. VP sin = V VP = V cosec
CIRCULAR MOTION
here cos =
3 x 1 3R = . = 5 5 R R
sin =
VP =
4 5
cosec =
5 V 4
5 4
...Ans.
gy (a)
tcNM+ABxfrdjsxh]fcUnq‘P’ges'kko`ÙkijfLFkrgksxkA fp=kkuqlkjbldkosx‘VP’o`Ùkdsvuqfn'kgksxkA;fnfcUnqPNM+‘AB’ijfLFkrgksxkrksbldk‘x’fn'kkesa?kVd ‘V’gksukpkfg,A
VP sin = V
3 x 1 3R = . = 5 5 R R
;gk¡ cos =
sin =
VP =
Sol. (b)
=
VP = V cosec
4 5
cosec =
5 V 4
5 4
...Ans.
VP 5V = R 4R
ALTERNATIVE SOLUTION : oSdfYid gy Sol. (a) Let ‘P’ have coordinate (x, y) ekuk‘P’dsfunsZ'kkad(x,y) x = R cos , y = R sin . dx d = – R sin = V dt dt
VX =
rFkkand
VY = R cos VP =
Sol. (b) = 97.
Vx2 Vy2 =
d = R cos dt
d V = dt R sin
V = – V cot R sin
V 2 V 2 cot 2 = V cosec ...Ans.
VP 5V = R 4R
A force of constant magnitude F acts on a particle moving in a plane such that it is perpendicular to the velocity v ( | v | = v) of the body, and the force is always directed towards a fixed point. Then the angle turned by the velocity vector of the particle as it covers a distance S is :(take mass of the particle as m) fu;r ifjek.k dk cy Ffdlh ry esa xfr dj jgs d.k ij bl rjg dk;Zjr gS fd ;g d.k ds osx v (| v | = v)ls yEcor~ gSrFkkbldhfn'kkfdlhfLFkjfcUnqdhrjQgSAd.kdsSnwjhr;djusijd.kdsosxlfn'k}kjkr;dks.kgksxk:(d.k dk nzO;eku mgS) [Made PKS, 2005] [M.Bank(07-08)_Circular Motion_1.7]
CIRCULAR MOTION
(A) Sol.
(B)
2
2FS
(C)
2
2mv mv Since F V , the particle will move along a circle.
98.
FS
F=
mv 2 R
& =
S R
=
FS2 mv
(D*)
FS mv 2
FS mv 2
A small bead of mass m is in equilibrium at the position shown on a smooth vertical ring of radius r. The ring revolves at some constant angular velocity about vertical diameter. Find: rf=kT;kdh,dm/okZ/kjfpduhoy;ijn'kkZ;hx;hfLFkfrijmnzO;ekudk,dNksVkNYyklkE;koLFkkesagSAoy;m/ okZ/kj O;kl ds çfr fdlh fu;r dks.kh; osx ls ?kwerh gSA Kkr dhft,A [M.Bank(07-08)_CircularMotion_2.26]
(i) (ii)
angular velocity of the ring. oy;dkdks.kh;osxA if m is displaced slightly from its equilibrium position, prove that it will execute S.H.M. on the ring. Find its time period. ;fnmdksbldhlkE;koLFkklsFkksM+kfoLFkkfirdjdsNksM+ktk;srksfl)dhft,fd;goy;ijl-vk-x-djsxkA
bldk vkorZdky Kkr dhft,A g (ii) r cos
[Ans: (i) 99.
2
r cos g sin 2
]
A small bead of mass m = 1 kg is carried by a circular hoop having centre at C and radius r = 1 m which rotates about a fixed vertical axis. The coefficient of friction between bead and hoop is µ = 0.5. The maximum angular speed of the hoop for which the bead does not have relative motion with respect to hoop. ,do`Ùkkdkjywiftldhf=kT;kr=1mrFkkdsUnzCgS];gfLFkjÅ/okZ/kjv{kdslkis{k?kw.kZudjjghgSAblijm= 1kgdhNksVhchMfLFkrgSAywirFkkchM¼eudk½dse/;?k"kZ.kxq.kkadµ=0.5gSAywidslkis{kchM¼euds½dhlkis{k
xfrughagksblds fy,ywidhvf/kdredks.kh;pkyD;kgksxh\
[M.Bank(07-08)_C.M._2.5]
[MB_Q. 2.5]
(A) 5 2 Sol.
1/ 2
(B) 10 2
1/ 2
(C) 15 2
1/ 2
[Made 2005, MPS]
(D*) 30 2
1/ 2
(D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is
CIRCULAR MOTION
For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N where N = mg cos 45° + m2 (r sin 45º) sin 45° From 1 and 2 we get. =
gy
30 2
.................... (1) .................... (2)
rad / s.
(D) tcchM Åij dh vksjBhd fQlyusdh fLFkfresa gksrkgS rc ywi dkdks.kh; osxvf/kdre gSAchM dk freebody diagram
chMdks Åijdh vksjugha fQlyusds fy, m2 (r sin 45°) cos 45° – mg sin 45° < N tgk¡ N = mg cos 45° + m2 (r sin 45º) sin 45° leh01 rFkk2 ls = 100.
30 2
rad / s.
A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the track at B is: [Made 2004] fp=kkuqlkjÅ/okZ/kjryesaRf=kT;kdkfpdukfLFkjo`ÙkkdkjiFkçnf'kZrgSA,dCykWddksfLFkfrAlsNksM+usij;giFk dksfcUnqBijNksM+rkgSrksfcUnqBijbldsiFkdhoØrkf=kT;kgS: [M.Bank(07-08)_C.M._7.2]
(A) R
Sol.
.................... (1) .................... (2)
(B)
A R–R cos53 =2R/5
R 53º
R 4
O 37º B
R–R cos37= R/5 Reference line 37º (funs ' k j s[ kk) Z g g cos37
By energy conservation between A & B A rFkkBijÅtkZlaj{k.kdsfu;els
(C*)
R 2
(D)noneof thesebuesalsdksbZugha
CIRCULAR MOTION
Mg
2R MgR 1 +0= + MV2 5 5 2
V=
2gR 5
V2 2gR / 5 R Now, radius of curvature r = ar g cos 37 2 V2
2gR / 5
R
oØrk f=kT;k r= a g cos 37 2 r
101.
A particle tied with a string moves in a vertical circular path. If X1 and X2 are the extreme left and right positions of the particle on the path, whereas Y1 and Y2 the extreme upper and lower positions, then the tension(s) in the string at the position(s) [REE_1990] [M.Bank(07-08)_Circular Motion_3.71] ,dd.kdksjLlhlsck¡/kdjm/okZ/kjo`Ùkh;xfrdjkrsgSA;fnX1vksj+X2 d.kdsiFkdsfy,ckW;hrFkknkW;hmPpre (extreme) fLFkfr rFkkY1 vksjY2 vf/kdre mPp (extremeupper) rFkk U;wure fLFkfr jLlh esaruko gS rks % (A) X1 is greater than that at X2 (B*) Y2 is greater than that at Y1 (C) Y1 and Y2 are the same (D) Y1, Y2, X1 and X2 are equal (A) X1 esaX2dhvis{kkvf/kdregksxkA (B*) Y2 esaY1dhvis{kkvf/kdregksxkA (C) Y1 rFkkY2 esa,d lekugksxkA (D) Y1, Y2, X1 , X2 esa leku gksxkA
102.
A ball attached with massless rope of the length swings in vertical circle as shown in figure. The total acceleration of the ball is a ˆi bˆj m/sec2 when it is at angle = 37º; (where a and b are positive constant). Find the magnitude of centripetal acceleration of the ball at the instant shown. The axis system is shown in figure.
,dxsandksyEckbZdhnzO;ekughujLlhlstksM+djÅ/oZo`Ùkesafp=kkuqlkj?kqek;ktkrkgSAtc;gÅ/oZls=37º dk dks.kcukjghgSrcbldkdqyRoj.k a ˆi bˆj m/sec2gS(tgk¡arFkkb/kukRedfu;rkadgS)bl{k.kijvfHkdsUnzh;Roj.k dkifjek.kcrkb,Av{kfudk;fp=kesan'kkZ;kx;kgSA [Made BKM - 2005] [M.Bank(0708)_CM_3.2]
Sol.
Normal or centripetal acceleration of the ball is component of net acceleration along PO.
Ans.
ac = a cos 53° + b cos 37° = 3a 4b 5 5
3a 4b + m/sec2 5 5
CIRCULAR MOTION
103.
A toy car revolves in a circular path of radius 1m on a horizontal rough plane. It’s speed varies as V = 2t2. It starts sliding at t = 1s. Find the value of coefficient of friction between ground and the wheels of the car. Use g = 10 m/s2. ,d f[kyksuk dkj r= 1m f=kT;k ds {ksfrtµ[kqjnjs ry esa o`fr; xfr dj ldrh gSA bldh pky V= 2t2 ls cnyrh gSA ;gt=1sijfQlyuk 'kq:gks tkrhgSA rks dkj dsifg;s rFkklrg ds chp ?k"kZ.k xq.kkd D;kgksxkA (g= 10m/s2). [Made 2005 RKV] [2]
Sol.
Centripetal acceleration after 1s ac = at = tangential acceleration = Total acceleration =
v2 (2 12 )2 = = 4 m/s2 R 1
dv = 4t = 4 m/s2 dt
a c 2 a t 2 = 4 2 m/s2
when sliding starts 4 2 = µg ; 104.
µ =
4 2 2 2 = 10 5
Ans. µ =
2 2 5
A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a position where string makes an angle of 300 with the horizontal. On the circular path of the bob in vertical plane there is a peg ‘B’ at a symmetrical position with respect to the position of release as shown in the figure. If Vc and Va be the minimum speeds in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg ‘B’ then ratio Vc : Va is equal to : ,djLlhds,dfljslsckWctqM+kgSrFkknwljkfljkpegAlstqM+kgSAckWcdks{kSfrtls300fLFkfrrdystk;ktkrk gSrFkk;gkalsNksM+ktkrkgSAÅ/okZ/kjckWcdso`ÙkkdkjiFkij peg‘B’fLFkrgSAvcckWcdks;gkalsNksM+ktkrkgSAnf{k.kkorZ rFkkokekorZfn'kkvksa lsckWcdsPegBijVdjkus dsfy,U;wureosxØe'k%Vc rFkkVa gksrksVc :VagS:[OldRRB, Q. 83] [M.Bank(07-08)_C.M._3.68]
[Made BKM,2006, GRSTUX]
(A) 1 : 1 Ans.
(B) 1 :
(C*) 1 : 2
2
(D) 1 : 4
(C) For anti-clockwise motion, speed at the highest point should be
gR . Conserving energy at (1) & (2) :
(C)okekorZfn'kk esaxfrdsfy,] mPprefcUnq ijpky gR gksuhpkfg,A (1)rFkk(2)dse/;ÅtkZlaj{k.kls:
1 R 1 mv a2 = mg m(gR) 2 2 2
v a2 = gR + gR = 2gR
va =
2gR
For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not become loose any where until it reaches the peg B. nf{k.kkorZxfrdsfy,ckWcdsiklçkjEHkesadelsdebrukosxgksukpkfg,ftllsoks
tk;sA At the initial position :
çkjfEHkdfLFkfresa
CIRCULAR MOTION
T + mgcos600 =
mv c2 R
;
VC being the initial speed in clockwise direction. VCdhçkjfEHkdpkynf{k.korZfn'kkesagSA
105.
Sol.
For
VC min : Put T = 0 ; VC min : ds fy, T = 0 j[kh gSA
VC =
gR 2
gR 2 = 1 2 2gR
VC/Va =
VC : V a = 1 : 2
Ans.
A 10kg ball attached at the end of a rigid rod of length 1m rotates at constant speed in a horizontal circle of radius 0.5m and period 1.57 s as shown in the figure. The force exerted by the rod on the ball is 1myEchn`<+NM+ds,dfljsijca/kh10kgdhxsan 0.5mf=kT;kds{kSfrto`Ùkesfu;rosxls?kwerhgSA¼fp=kkuqlkj½ bldk vkorZdky 1.57sgSA NM+ }kjk xsan ij yxk;k x;k cy gksxk \ [M.Bank(07-08)_C.M._2.22]
(A) 158 N (B*) 128 N Let F be the force exerted by rod on the ball ; F cos = m2r F sin = mg F2 (mg)2 + (m2r)2
(D) 98 N (g = 10 ms-2)
(C) 110 N
Fv F
F=m
g 2 ( 2 r ) 2
2 = 4 rad/sec. 1.57 m = 10 kg and r = 0.5 m F = 128 N.
put =
106.
FH
mg
Two particles P and Q start their journey simultaneously from point A. P moves along a smooth horizontal wire AB. Q moves along a curved smooth track. Q has sufficient velocity at A to reach B always remaining in contact with the curved track. At A, the horizontal component of velocity of Q is same as the velocity of P along the wire. The plane of motion is vertical. If t1, t2, are times taken by P & Q respectively to reach B then (Assume velocity of P is constant) nksd.kPoQviuh;k=kk,dlkFkfcUnqAlsizkjEHkdjrsgSaA P,dfpdus{kSfrtrkjABdsvuqfn'kxfrdjrkgSA Q ,d fpdusoØh; iFkds vuqfn'kxfr djrkgSAoØh; iFkds lEidZesajgrsgq;sBrdig¡qpus ds fy;sQ dsiklA iji;kZIrosxgSAAijQdsosxdk{kSfrt?kVdrkjdsvuqfn'kPdsosxdscjkcjgSAxfrdkryÅ/okZ/kjgSA;fn P o Q }kjk Brd ig¡qpus esa fy;s x;s le; Øe'k% t1o t2 gks rks (Pdk osx fu;r ekusa)& [M.Bank(07-08)_CM_2.28]
CIRCULAR MOTION
(A) t1 = t2
(B*) t1> t2
(C) t1 < t2
(D)noneof thesebuesalsdksbZugha
V Sol.
Nx N
N A y B The horizontal component of velocity of Q will increase and become maximum at the top ; and will again become same at B. Because of its greater horizontal velocity the particle Q will reach B earlier than P t1 > t2 . 107.
A particle inside a hollow sphere of radius r, having a coefficient of friction (1/3) can rest up to a height of _______. [M.Bank(07-08)_Friction_1.22] rf=kT;kds[kks[kysxksysesa,dd.kfLFkrgSAxksysdhlrgdk?k"kZ.kxq.kkad(1/3)gSrksd.kfdruhšpkbZrdfojke esa jg ldrk gS _______ 3 [ Ans: r 1 2 ]
Sol.
mg sin = mg cos = 30º
tan =
3 Also h = R (1 – cos ) = R 1 – 2 Ans. 108.
A ball of mass 5 kg is connected to a pole by two strings of equal length L = 2 m. The pole rotates so that the ball moves in a circle with each string making an angle = 53o with vertical. If the period of the balls m circular motion is 2 sec. Find tension in each string. (sin 53o = 0.8 & 2 = 10) 5kgnzO;ekudh,dxsandkslekuyEckbZL=2mdhnksjkf'k;ksads}kjkiksy(pole)lstksM+ktkrkgSAiksyo`Ùkesaxfr djjgkgSAftllsxsançR;sdjLlhls=53o(Å/okZ?kjls)dks.kdslkFko`Ùkh;xfrdjrhgSA;fnxsandko`Ùkh;iFkesa vkorZdky 2 sec. gks rks çR;sd jLlh esa ruko Kkr djksA(sin 53o = 0.8 & 2 = 10) [DCM] [M.Bank(07-08)_C.M._5.12]
[ Ans: T 1 = 275/3 N, T 2 = 25/3N ] 109.
A particle tied to a string of length l is given a velocity at lowest point which is insufficient to complete the circular path in the vertical plane. The other end of the string is fixed. The radius of curvature of the path just after the string slacks is: [M_Bank(07-08)_C.M._7.13] ,dyEckbZdhjLlhls,dd.kcka/kktkrkgSrFkkbldksfuEurefcUnqlsdqNosxnsdjNksM+rstksfdiw.kZpØdsfy;s
vi;kZIrgSAjLlhdknwljkfljkfLFkjgSAjLlhds
(A*) (B) 2 (C) /2 As the particle is in circular path upto that point, radius of curvature in .
(D) 3
CIRCULAR MOTION
110.
A ball suspended by a thread swings on a vertical plane so that its acceleration in the extreme position and lowest position are equal. Angle of thread deflection in the extreme position will be: jLlhlstqM+hgqbZ,dxsanÅ/okZ/kj ryesa?kwejghgSAmfPp"BfcUnq(extremeposition)rFkkfuEfu"BfcUnq(lowest position) ij Roj.k cjkcj gSA mfPp"B fLFkfr esa jLlh dk fopyu dks.k (threaddeflection) gksxkA [M.Bank(07-08)_CM_3.34]
(A*) 2 tan1 Sol.
1 2
(B) tan1
1 2
(C) tan1 2
(D) tan1 2
aA = g sin(only tangential) aB =
v2 (only radial)
K.E. + P.E. = K.E. + P.E. =
1 1 m0 2 mg(1 cos ) mv 2 2 2 v 2 = 2g (1 – cos) ............(i)
aB =
v
A
B mgsin
mg
mgcos
v2 = 2g(1 – cos)
Since, aA = aB g sin = 2g(1 – cos)
111.
2sin
tan
cos = 2 × 2sin2 2 2 2
1 = 2 2
1 = 2 tan–1 2
Ans.(A)
A right circular cone is fixed with its axis vertical and vertex down. A particle is in contact with its smooth inside surface and describes circular motion in a horizontal plane at a height of 20 cm above the vertex. Find its velocity in m/s. ,dledks.ko`Ùkh;'kadq(rightcircularcone) ftldhv{km/okZ/kjrFkkftldk'kh"kZuhpsgSafLFkj¼tM+or½gSA,dd.k tksbldhfpduhvkUrfjd lrgij'kh"kZls20cmÅij{kSfrt ryesao`Ùkh;xfr djjgkgSrksbldkosxgksxk
[M.Bank(07-08)_Circular Motion_5.3]
[ Ans: 2 m/s ] 112.
A solid sphere is placed on a smooth horizontal surface. A sudden blow is given horizontally to the sphere at a height h = 4R/5 above the centre line. The minimum time after which the highest point B will touch the ground is _______, if I is the impulse of the blow. The displacement of the centre of mass during this interval is _______. ,dBkslxksykfpduh{kSfrtlrgijj[kktkrkgSAdsUnzlsh=4R/5Åpk¡bZijvpkudxksysdks{kSfrtfn'kkesavkosx
fn;ktkrkgSAU;wurele;_______gksxktcxksysdkmPprefcUnqBtehuijigq¡prkgSA;fnvkosxgSAblçfrfØ;k esanzO;ekudsUnzdkfoLFkkiu_______ gksxkA [M_Bank (07-08)_Rotation_6.26] Sol.
Using linear impulse momentum equation
CIRCULAR MOTION
M Using angular impulse momentum equation wrt centre
= mv
.
v=
4 2 R= MR2 5 5
2 .............(2) MR Point B has to traverse angle to reach ground. As is constt., time required
=
MR = MR = 2 2 In the same time ‘t’ COM displacement
t=
mR R . = m 2 2 A circular road of radius R is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circular road, the friction co-efficient between the tyre & road is negligible: [6.6_Circular Motion] (A) the car cannot make a turn without skidding (B) if the car runs at a speed less than 40 km/hr, it will slip up the slope (C) if the car runs at the correct speed of 40 km/hr, the force by the road on the car is equal to mv 2/r (D*) if the car runs at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv 2/r
= v.t =
113.
114.
A ball tied to the end of a string swings in a vertical circle under the influence of gravity (A) when the string makes an angle 90º with the vertical , the tangential acceleration is zero & radial acceleration is somewhere between maximum and minimum (B*) when the string makes an angle 90º with the vertical, the tangential acceleration is maximum & radial acceleration is somewhere between maximum and minimum (C) at no place in the circular motion, tangential acceleration is equal to radial acceleration (D*) throughout the path whenever radial acceleration has its extreme value, the tangential acceleration is zero. [3.56_Circular Motion]
115.
A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s2 in a circular path. If it slips when its total acceleration becomes 1 m/s2, then the angle through which it would have turned before it starts to slip is : [Made CSS 2006, GRST] [M.Bank(07-08)_C.M._2.32] ,dd.k,do`ÙkkdkjiFkesa0.6m/sec2dsfu;rLi'kZjs[kh;Roj.klsxfrdjukizkjEHkdjrkgSAtcbldkdqyRoj.k 1m /sec2 gkstkrkgSarks;gfQlytkrkgSAfQlyukizkjEHkdjuslsBhdigysrdblds}kjk?kwekgqvkdks.kgksxk
& (A) 1/3 rad Sol.
aNet =
so
(C) 4/3 rad
(D) 2 rad
a 2t a 2c
2 = 2 + 2 = 0 2 = 2 2R = 2 (R) ac = 2R = 2at 1=
(B*) 2/3 rad
0.36 (1.2 )2
1 – 0.36 = (1.2 )2
0.8 1.2
2 radian 3
CIRCULAR MOTION
116.
Sol.
A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle, with a uniform angular speed of 2 rad/sec. Then the tension in the string will be : 1kgnzO;ekuds,diRFkjdks1myEckbZdhjLlhds,dfljslsck¡/kktkrkgStks{kSfrto`Ùkesa,dlekudks.kh;pky 2rad/sec.ls?kqejghgSrksjLlhesarukogksxk& KCET_1998 [M.Bank(07-08)_C.M._2.33] (A*) 4 N (B) 8 N (C) 1 N (D) 2 N Given : Mass of the stone m = 1 kg Radius of the circle r = 1 m Angular velocity 2 rad / sec The relation for the tension is given by m 2 r 1 22 1
117.
4N
A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60o with the vertical. Then : (g = 9.8 m/s2) ,dHkkjhd.k1.6myEckbZdhjLlhds,dfljsijtksM+ktkrkgSrFkknwljkfljkOtM+Ro¼fLFkj½gSA;g,d'kadqnksyd (conical pendulum) dh rjg xfr djrk gSA ftldk Å/okZ/kj ls dks.k 60o gSA (g = 9.8 m/s2) [M.Bank(07-08)_C.M._1.50] (A*) its period of revolution is
4 sec. 7
bldk vkorZdky
4 sec. lSd.M gksxkA 7
(B*) the tension in the string is doubled the weight of the particle
jLlhesarukod.kdsHkkjdknqxqukgksxkA (C*) the velocity of the particle = 2.8
3 m/s
d.k dk osx = 2.8 3 m/sgksxkA
(D*) the centripetal acceleration of the particle is 9.83 m/s2. d.kdkvfHkdsUnzh;Roj.k9.83m/s2 gksxkA T cos 60o
60o 60o
/ 2
T 3/2
V
Sol.
o
T sin 60
mg
2
mv T 3 = ( 3 / 2) 2
T = mg 2 Hence T = 2 mg , So (B) holds From (1) & (2) V2 = 3 g/2 3 9 .8 1 .6 2
V=
V = 2.8
3 m/s2 . So (C) hold
( 3 g / 2)
ac = V2/r =
( 3 / 2)
(D) holds.
=
3g
........(1)
.......(2)
CIRCULAR MOTION
t=
2 3 / 2 2 r = (3 g / 2) v
=
2g
2
t = 4/7 (A) holds. 118.
A weightless rod of length 2 carries two equal masses 'm', one secured at lower end A and the other at the middle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C. What horizontal velocity must be imparted to the mass at A so that it just completes the vertical circle. 2yEckbZdhnzO;ekujfgrNM+lsnks'm'nzO;ekudsd.kNM+dsfuEureArFkke/;fcUnqBijtqM+sgq,gSANM+Å/okZ/ kjryesafLFkrfcUnqClsxqtjusokyh{kSfrtv{klsxqtjrhgS?kqek;ktkrkgSAAfcUnqijfLFkrnzO;ekudksfdrus{kSfrt osx fn;k tk,rkfd ;gBhd Å/okZ/kjo`Ùk r; dj ldsA [Made MPS - 2005] [ M . Ban k( 0 708)_C.M._3.5]
Sol.
Let the initial velocity given to the mass at A be u. Then the velocity of mass at B is u/2 As the system moves from initial the final position Increase in potential energy is = 4 mg + 2mg 2
1 1 u 5 2 Decrease in kinetic energy = mu m = mu2 2 2 2 8
From conservation of energy 5 mu2 = 6 mgl 8
or
u=
48 g 5
Comprehension (119 - 121) vuqPNsn [M_Bank(07-08)_Circular Motion_2.37( 1,2,3)] Two cars A and B start racing at the same time on a flat race track which consists of two straight sections each of length 100 and one circular section as in fig. The rule of the race is that each car must travel at constant speed at all times without ever skidding Modifed VSS nksdkjAoB,dghle;ij,dleryiFkijnkSM+ukizkjEHkdjrhgSaAbliFkesankslh/ksHkkxizR;sddhyEckbZ100 o,do`ÙkkdkjHkkxfp=kkuqlkjgSAnkSM+dkfu;e;ggSfdnksuksadkjiwjsle;fcukfQlysfu;rpkylsnkSM+rhgSaA rB rA
rA= 100 m rB=200 m L = 100
B A
CIRCULAR MOTION
119.
If µA = 0.1, µB = 0.2 (A is coefficient of friction on track A and B is the coefficient of friction on track B) then : ;fnµA=0.1,µB=0.2rks(tgkaA AjkLrsij?k"kZ.kxq.kkad,oaB,BjkLrsij?k"kZ.kxq.kkadgS): (A) car A completes its journey before car B (B*) both cars complete their journey in same time on circular part (C*) speed of car B is greater than that of car A
(D*) car B completes its journey before car A.
(A) dkj Abldh ;k=kk dkj Blsigys iwjh djrh gSA (B*)nksuksadkjo`ÙkkdkjiFkijmudh;k=kklekule;esaiwjhdjrhgSaA (C*) dkjB dh pky dkj Adh pky ls vf/kd gSA Sol.
(Vmax)A =
A rA g =
0.1 100 10 = 10 m/s
(Vmax)B =
B rB g =
0.2 200 10 = 20 m/s
(D*)dkj Bbldh ;k=kk dkj Als igys iwjh djrh gSA
VB > VA Ans. Journey time on circular part rA 100 tA = V = = 10 sec 10 A rB 200 tB = V = = 10 sec 20 B
tA = tB Ans. Total journey time TA =
100 100 rA 300 = = 30 sec VA 10
TB =
100 100 rB 400 = = 20 sec VB 20
TB < TA Ans. 120.
If speed of car A is 108 kmph and speed of car B is 180 kmph, and both tracks are sufficiently rough : ;fn dkjAdh pky108kmphodkj Bdhpky 180kmphgS,oa nksuksaiFk i;kZIr :i ls [kqjnjs gSaA (A) car A completes its journey before car B (C) speed of car A is greater than that of car B
Sol.
(B) both cars complete their journey in same time (D*) car B completes its journey before car A.
(A) dkjAbldh;k=kk dkj Bls igys iwjh djrh gSA
(B)nksuksadkjmudh;k=kklekule;esaiwjhdjrhgSA
(C) dkjAdh pky dkj B dh pky ls vf/kd gSA
(D*)dkj Bbldh ;k=kk dkj Als igys iwjh djrh gSA
VA = 108 ×
5 = 30 m/s 18
VB = 180 ×
5 = 50 m/s 18
Total Journey Time TA =
100 100 rA 300 = = 10 sec VA 30
TB =
100 100 rB 400 = = 8 sec VB 50
TB < TA Ans. 121.
If VB = 90kmph, the minimum value of µA so that car A can complete its journey before car B is :
CIRCULAR MOTION
;fnVB = 90kmph, µAdk U;wure eku rkfd dkj Abldh ;k=kk dkj B ls igys iwjh dj ldsA (A*)
Sol. 122.
45 128
(B)
45 100
(C)
45 64
(D)Noneof thesebuesa ls dksbZ ugh
V2 45 µ= = rg 128
400 (m) 300 (m) 75 t= = 16 second ; v A = = m/s ; 25 (m / s) 16 (sec) 4
A disc of radius R has a light pole fixed perpendicular to the disc at the circumference which in turn has a pendulum of length R attached to its other end as shown in figure. The disc is rotated with a constant angular velocity . The string is making an angle 300 with the rod. Then the angular velocity of disc is: R f=kT;kdh pdrhdh ifjf/kij pdrhds yEcor,d gYdhNM+ tqM+h gS] ftldsnwljs fljslsRyEckbZdk ,dyksyd fp=kkuqlkjtqM+kgSApdrhdksfLFkjdks.kh;osxls?kqekrsgSaAjLlhNM+ls300dkdks.kcukrhgSArcpdrhdkdks.kh; osxgS: [M.Bank(07-08)_CM_5.2]
[Old RRB, Q. 84] [Made 2006, RS, GRSTUX]
1/ 2
3g (A) R
Ans.
1/ 2
1/ 2
3g (B) 2 R
g (C) 3 R
(D) The bob of the pendulum moves in a circle of radius (R + Rsin300) = (D)yksyd dk ckWc (R+ Rsin300)=
1/ 2
2g (D*) 3 3 R
3R 2
3R f=kT;k ds o`Ùkesa xfr djrk gSA 2
3R 2 Tsin300 = m 2 Tcos300 = mg
Force equations cy lehdj.ksa :
123.
tan300 =
1 3 2 R = 2 g 3
=
2g 3 3R
Ans.
A solid body at time t = 0 starts rotating about fixed axis with a time dependent angular acceleration given by = kt where k is constant. For an arbitrary point of the body, the time taken for its total acceleration vector to be at angle with its linear velocity vector is _______. ,dBksloLrq le;t= 0ij,d fLFkjv{k dsifjr%le;ijfuHkZjdks.kh; Roj.k=ktdslkFk ?kweukçkjEHk djrhgSA tgkakfu;rkadgSAoLrqds,dLoSfPNdfcUnqdsfy,bldsdqyRoj.klfn'kdksbldsjs[kh;osxlfn'kdslkFkdks.k cukus esa yxk le; gS _______ [4 min.] [M.Bank(07-08)_CM_1.28]
CIRCULAR MOTION
1
Sol.
4 tan 3 [ Ans: ] k = kt at = r = kt
=
a a t ac
d = kt dt t
d ktdt
0
=
0
kt 2 k 2t4 r , aC = 2r = 2 4
ac k 2 t 4r / 4 kt 3 tan = = = at ktr 4
124.
1/ 3
4 tan t= k
Two particles P and Q start their journey simultaneously from point A. P moves along a smooth horizontal wire AB. Q moves along a curved smooth track. Q has sufficient velocity at A to reach B always remaining in contact with the curved track. At A, the horizontal component of velocity of Q is same as the velocity of P along the wire. The plane of motion is vertical. If t1, t2, are times taken by P & Q respectively to reach B then (Assume velocity of P is constant) nksd.kPoQviuh;k=kk,dlkFkfcUnqAlsizkjEHkdjrsgSaA P,dfpdus{kSfrtrkjABdsvuqfn'kxfrdjrkgSA Q ,d fpdusoØh; iFkds vuqfn'kxfr djrkgSAoØh; iFkds lEidZesajgrsgq;sBrdig¡qpus ds fy;sQ dsiklA iji;kZIrosxgSAAijQdsosxdk{kSfrt?kVdrkjdsvuqfn'kPdsosxdscjkcjgSAxfrdkryÅ/okZ/kjgSA;fn P o Q }kjk Brd ig¡qpus esa fy;s x;s le; Øe'k% t1o t2 gks rks (Pdk osx fu;r ekusa)& [M.Bank(07-08)_CM_2.28]
(A) t1 = t2
(B*) t1> t2
(C) t1 < t2
(D)noneof thesebuesalsdksbZugha
V
Sol. A
Ny
Nx N B
The horizontal component of velocity of Q will increase and become maximum at the top ; and will again become same at B. Because of its greater horizontal velocity the particle Q will reach B earlier than P t1 > t2 . 125.
A ring of radius R lies in vertical plane. A bead of mass ‘m’ can move along the ring without friction. Initially the bead is at rest at the bottom most point on ring. The minimum horizontal speed v with which the ring must be pulled such that the bead completes the vertical circle [M.Bank(07-08)_Circular Motion_3.9] R f=kT;kdkoy;m/okZ/kjryesafLFkrgSA‘m’nzO;ekudk,deudkoy;dsvuqfn'kfcuk?k"kZ.kdsxfrdjldrkgSA
izkjEHkesaeudkoy;dsU;wurefcUnqijfojkeesagSAoy;dksfdlU;wurepkyls[khapukpkfg,ftllseudko`Ùkh; xfr dj ldsA
CIRCULAR MOTION
[Made MPS-2005]
(A) Sol.
(B*)
3gR
(C)
4gR
5gR
(D)
5 . 5 gR
In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is v at the lowest position.
The condition for bead to complete the vertical circle is, its speed at top position v top 0 From conservation of energy 1 1 m v 2top + mg (2R) = mv 2 2 2
or 126.
v=
4 gR
A right circular cone is fixed with its axis vertical and vertex down. A particle is in contact with its smooth inside surface and describes circular motion in a horizontal plane at a height of 20 cm above the vertex. Find its velocity in m/s. (Take g = 10 m/s2) ,dledks.ko`Ùkh;'kadq(rightcircularcone) ftldhv{km/okZ/kjrFkkftldk'kh"kZuhpsgSafLFkj¼tM+or½gSA,dd.k tksbldhfpduhvkUrfjdlrgij'kh"kZls20cmÅij{kSfrtryesao`Ùkh;xfrdjjgkgSrksbldkosxgksxkA(g=10 m/s2)
[M.Bank(07-08)_Circular Motion_5.3]
[ Ans: 2 m/s ] Sol.
mgcos =
mv 2 sin r N mV 2 r
mg r = tan n
127.
v=
gh =
10
r = htan
20 = 100
2 m/s.
Two blocks A and B each of same mass are attached by a thin inextensible string through an ideal pulley. Initially block B is held in position as shown in figure. Now the block B is released. Bolck A will slide to right and hit the pulley in time tA. Block B will swing and hit the surface in time tB. Assume the surface as frictionless. [M_Bank_Circular_Motion _Q. 3.33] nkslekunzO;ekudsCykWdArFkkBvfoLrkfjronzO;ekujfgrjLlhds}kjkvkn'kZf?kjuhlstksM+stkrsgSaizkjEHkesaCykWd Bdksfp=kkuqlkjj[kktkrkgSAvcCykWdBdksNksM+ktkrkgSACykWdAlrgijfQlyrkgqvkf?kjuhlstAle;esaVdjkrk
CIRCULAR MOTION
gSACykWdB>qyrkgqvklrglstBle;esaVdjkrkgSA¼ekuklrg?k"kZ.kjfgrgSA½
Sol.
(A) tA = tB (B*) tA < tB (C) tA > tB (D) data are not sufficient to get relationship between tA and tB. lwpukvi;kZIrgSblfy,tArFkktBesalaca/kiznf'kZrughafd;ktkldrkA At any instant external force in horizontal direction is T for A and Tcos for B and T is always greater then Tcos.
T A
T Tcos B
128.
A stone of mass M is tied at the end of a string, is moving in a circle of radius R, with a constant angular velocity . The total work done on the stone, in any half circle, is : M.Bank_CM_5.11 [GRSTU CT-2 (2309)_Paper-2_Q.4]
jLlhds,dfljsijnzO;ekuMdsiRFkjdksck¡/kdj]f=kT;kRdso`Ùkesafu;rdks.kh;osxls?kwek;ktkrkgSaArksfdlh Hkhvk/kso`ÙkesaiRFkjdks/kqekusesafd;kx;kdqydk;Zgksxk& (A) MR2 2 (B) 2 MR2 2 (C) MR2 2 (D*) 0 Sol.
(Easy)
Since there is no change in kinetic energy of stone, the total work done on stone in any duration is
zero. (Easy)pwafdiRFkjdhxfrtÅtkZesadksbZifjorZuughagksrkgSvr%fdlhHkhle;kUrjkyesaiRFkjijfd;kx;kdk;Z'kwU;
gksxkA 129.
A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating and is fp=kkuqlkj'r'f=kT;kdsfpdusfLFkjv)ZxksysdsfcUnqAls,dd.kfojkekoLFkklsxfrçkjEHkdjrkgSAd.kdkBfcUnq ij lEidZNwV tkrkgSACv)ZxksysdkdsUnzgSrksrFkk dkstksM+us okyhlehdj.kgS& [MB_Q.3.10] [M.Bank(07-08)_Circular Motion_3.10]
[Made 2005, MPS]
(A) 3 sin = 2 cos Sol.
(B) 2 sin = 3 cos
(C*) 3 sin = 2 cos
(D) 2 sin = 3 cos
(C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv 2 = mg sin .......... (1) r Applying conservation of energy as the block moves from A to B..
1 mv 2 = mg (r cos – r sin ) 2 Solving 1 and 2 we get 3 sin = 2 cos
.......... (2)
CIRCULAR MOTION
gy
(C) d.k dk lEidZ NwVus ls rqjUr igys d.k dk Bij osx dkVgSA
xksyh; lrg ds yEcor~ d.k ij U;wVu dk fu;e yxkus ij mv 2 = mg sin .......... (1) r CykWddks AlsBrdxfr djusij ÅtkZlaj{k.k dsfu;e ls
1 mv 2 = mg (r cos – r sin ) 2
.......... (2)
1 rFkk 2dks gy djus ij 3 sin = 2 cos 130.
A wet open umbrella is held upright and whirled about the handle with a uniform angular speed of 21 revolutions in 44 sec. If the rim of the umbrella is a circle of diameter 1m, and height of the rim from the ground is 1.5 m, find the radius of the circle along which the drops of water spun off the rim tangentially without any relative velocity hit the ground. ,dcjlkrhNkrsdksÅijdhvksjlh/ks[kM+kdjdsbldsgRFksdslkis{kbldks,dlekudks.kh;pky21?kw.kZu44 lSd.M ls?kqek;ktkrkgSA;fnNkrsdhifjf/k,d1mO;kldso`Ùkds:iesagksoifjf/kdh/kjkrylsÅpk¡bZ1.5mrksikuh dhcwankslscuso`Ùkdhf=kT;kD;kgksxhtksifjf/klsnwjgksusij/kjkryijfxjrhgSA [M.Bank(07-08)_C.M._1.27]
[Ans: Sol.
37 m] 40
From projectile motion 2h g
t= V h = 1.5
=V
2h g
=
21x 2 =3 44
r=
1 , 2
V = r =
=
R = radius =
=
3 2
3 10
1 2 2
9 x3 1 4 x 10 4
2
3 2
CIRCULAR MOTION
= 131.
37 m 40
A large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizontal circular path. If is the length of the string from m to the top end of the tube and is angle between this part and vertical part of the string as shown in the figure, then time taken by m to complete one circle is equal to [Made A.K.S. sir] ,dHkkjhnzO;ekuM gYdhjLlhlsfLFkj:ilsyVdjgkgSrFkkjLlhdknwljkfljkfpduhtM+or~V~;wclsxqtkjus dscknNksVsnzO;ekumlsca/kkgSAtks{kSfrto`Ùkh;iFkesa?kwejgkgSA;fnmlsV~;wcdsÅijhfljsrdjLlhdhyEckbZ vkSjjLlhdsblHkkxrFkkjLlhdsÅ/okZ/kjHkkxdschpdks.k gSrksm }kjk,dpDdjyxkusesao`Ùkiwjkdjus esafy;sx;sle;dkekugksxk: [M.Bank(07-08)_Circular Motion_2.29]
m M (A) 2 g sin
Sol.
(B) 2 g cos
For M to be stationary MfLFkjkoLFkkesagS& T = Mg Also for mass m, m dsfy, T cos = mg T sin =
m (C) 2 g M sin
m (D*) 2 g M
.... (1)
.... (2)
mv 2 .... (3) sin
dividing (3) by (2) (3)esa(2)dkHkkxnsusij
v2 tan = v= g sin
g . sin cos
2 sin Time period vkorZ dky
2 R = = v
g . sin cos
m From (1) and (2) cos = M lehdj.k (1)vkSj (2)ls
Tcos
m T M
Tsin
mg
Mg
m gM A particle starts from rest at O and moves along a horizontal semi circular track OAB of radius R = 1m as shown in the figure. The rate of change of speed of the particle is constant and equals to 2m/s2. A is a point lying exactly on the middle of semicircular track as shown in figure. When the particle reaches A. Find ,dd.kOlsfojkelsxfrdjrsgq,v/kZo`ÙkkdkjiFkOAB(f=kT;kR =1m)esafp=kkuqlkjxfrdjrkgSAd.kdhpky esaifjorZudhnjfu;ro2m/s2 gSAiFkdse/;esafcUnqAgSAtcd.kAijigq¡prkgSrks [M.Bank(07-08)_Circular Motion_1.4]
then time period vr% vkorZ dky = 2 132.
CIRCULAR MOTION
(a)
The magnitude of velocity vector of the particle at the instant.
bl{k.kd.kdsosxlfn'kdkifjek.kgksxkA (b)
Magnitude of acceleration vector of the particle at the instant.
bl{k.kd.kdsRoj.klfn'kdkifjek.kgksxkA (c)
The cosine of angle between acceleration and velocity vector of the particle at that instant bl{k.kRoj.klfn'krFkkosxlfn'kdschpdks.kgksrkscos gksxkA
[Made MPS - 2005]
Sol.
(a) The speed of particle at A is v 2 = u2 + 2as
u = 0, a = 2, s = 2 2
= 2
meters 2
v=
v = 2 ˆi m/s The normal acceleration of the particle at A is
(b)
2
v2 = 2m/s2 R tangential acceleration of the particle is = 2m/s2 Total acceleration vector is a = 2ˆi 2 ˆj m/s2 Magnitude of total acceleration is a = 4 4 2 = 2 1 2 m/s2 1 av (c) cos = = a v 1 2 133.
A force of constant magnitude F acts on a particle moving in a plane such that it is perpendicular to the velocity v ( | v | = v) of the body, and the force is always directed towards a fixed point. Then the angle turned by the velocity vector of the particle as it covers a distance S is :(take mass of the particle as m) fu;r ifjek.k dk cy Ffdlh ry esa xfr dj jgs d.k ij bl rjg dk;Zjr gS fd ;g d.k ds osx v (| v | = v)ls yEcor~
gSrFkkbldhfn'kkfdlhfLFkjfcUnqdhrjQgSAd.kdsSnwjhr;djusijd.kdsosxlfn'k}kjkr;dks.kgksxk:(d.k dk nzO;eku mgS) [Made PKS, 2005] [M.Bank(07-08)_Circular Motion_1.7] (A) Sol.
2
(B)
2FS 2
(C)
2mv mv Since F V , the particle will move along a circle.
FS2 mv
(D*)
FS mv 2
FS S mv 2 & = = R R mv 2 A rod AB is moving on a fixed circle of radius R with constant velocity ‘v’ as shown in figure. P is the point of
134.
FS
F=
3R from centre of the circle. 5 The velocity of the rod is perpendicular to the rod and the rod is always parallel to the diameter CD. fp=kkuqlkj,dNM+AB,Rf=kT;kdso`Ùkijfu;rosx‘v’lsxfrdjjgkgSAo`ÙkrFkkNM+dkçfrPNsnufcUnqPgSAfdlh 3R {k.ko`ÙkdsdsUnzlsNM+dhnwjhx= gSANM+dkosxNM+dsyEcor~gSrFkkNM+ges'kkO;klCDdslekUrjgSA 5 [M.Bank(07-08)_Circular Motion_1.5]
intersection of the rod and the circle. At an instant the rod is at a distance x =
CIRCULAR MOTION
(a) Find the speed of point of intersection P. çfrPNsnu fcUnqPdh pkyKkr djksA [MB_Q. 1.29] [Made MPS 2005] (b) Find the angular speed of point of intersection P with respect to centre of the circle. çfrPNsnufcUnqPdho`ÙkdsdsUnzdslkis{kdks.kh;pkyKkrdjksA [6]
Sol. (a)
As a rod AB moves, the point ‘P’ will always lie on the circle. its velocity will be along the circle as shown by ‘VP’ in the figure. If the point P has to lie on the rod ‘AB’ also then it should have component in ‘x’ direction as ‘V’. VP sin = V VP = V cosec here cos =
3 x 1 3R = . = 5 5 R R
sin =
VP =
4 5
cosec =
5 V 4
5 4
...Ans.
gy (a)
tcNM+ABxfrdjsxh]fcUnq‘P’ges'kko`ÙkijfLFkrgksxkA fp=kkuqlkjbldkosx‘VP’o`Ùkdsvuqfn'kgksxkA;fnfcUnqPNM+‘AB’ijfLFkrgksxkrksbldk‘x’fn'kkesa?kVd ‘V’gksukpkfg,A VP sin = V VP = V cosec ;gk¡ cos =
Sol. (b)
x 1 3R 3 = . = 5 5 R R
sin =
VP =
=
4 5
5 V 4 VP 5V = R 4R
cosec =
5 4
...Ans.
CIRCULAR MOTION
ALTERNATIVE SOLUTION : oSdfYid gy Sol. (a) Let ‘P’ have coordinate (x, y) ekuk‘P’dsfunsZ'kkad(x,y) x = R cos , y = R sin . dx d = – R sin = V dt dt
VX =
rFkkand
VY = R cos VP =
Sol. (b) = 135.
Vx2 Vy2 =
d = R cos dt
d V = dt R sin
V = – V cot R sin
V 2 V 2 cot 2 = V cosec ...Ans.
VP 5V = R 4R
A small bead of mass m = 1 kg is carried by a circular hoop having centre at C and radius r = 1 m which rotates about a fixed vertical axis. The coefficient of friction between bead and hoop is µ = 0.5. The maximum angular speed of the hoop for which the bead does not have relative motion with respect to hoop. ,do`Ùkkdkjywiftldhf=kT;kr=1mrFkkdsUnzCgS];gfLFkjÅ/okZ/kjv{kdslkis{k?kw.kZudjjghgSAblijm= 1kgdhNksVhchMfLFkrgSAywirFkkchM¼eudk½dse/;?k"kZ.kxq.kkadµ=0.5gSAywidslkis{kchM¼euds½dhlkis{k xfrughagksblds fy,ywidhvf/kdredks.kh;pkyD;kgksxh\ [M.Bank(07-08)_C.M._2.5]
[MB_Q. 2.5]
(A) 5 2 Sol.
1/ 2
(B) 10 2
1/ 2
1/ 2
(D*) 30 2
1/ 2
(D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is
For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N where N = mg cos 45° + m2 (r sin 45º) sin 45° From 1 and 2 we get. = 136.
(C) 15 2
[Made 2005, MPS]
30 2
.................... (1) .................... (2)
rad/s.
A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the track at B is: [Made 2004] fp=kkuqlkjÅ/okZ/kjryesaRf=kT;kdkfpdukfLFkjo`ÙkkdkjiFkçnf'kZrgSA,dCykWddksfLFkfrAlsNksM+usij;giFk dksfcUnqBijNksM+rkgSrksfcUnqBijbldsiFkdhoØrkf=kT;kgS: [M.Bank(07-08)_C.M._7.2]
CIRCULAR MOTION
(A) R
R 53º
A R–R cos53 =2R/5
Sol.
R 4
(B)
(C*)
R 2
(D)noneof thesebuesalsdksbZugha
O 37º B
R–R cos37= R/5 Reference line 37º (funs ' k j s[ kk) Z g g cos37
By energy conservation between A & B
Mg
V=
2R MgR 1 +0= + MV2 5 5 2
2gR 5
Now, radius of curvature r = 137.
V2 2gR / 5 R ar g cos 37 2
A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a position where string makes an angle of 300 with the horizontal. On the circular path of the bob in vertical plane there is a peg ‘B’ at a symmetrical position with respect to the position of release as shown in the figure. If Vc and Va be the minimum speeds in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg ‘B’ then ratio Vc : Va is equal to : ,djLlhds,dfljslsckWctqM+kgSrFkknwljkfljkpegAlstqM+kgSAckWcdks{kSfrtls300fLFkfrrdystk;ktkrk gSrFkk;gkalsNksM+ktkrkgSAÅ/okZ/kjckWcdso`ÙkkdkjiFkij peg‘B’fLFkrgSAvcckWcdks;gkalsNksM+ktkrkgSAnf{k.kkorZ rFkkokekorZfn'kkvksa lsckWcdsPegBijVdjkus dsfy,U;wureosxØe'k%Vc rFkkVa gksrksVc :VagS:[OldRRB, Q. 83] [M.Bank(07-08)_C.M._3.68]
[Made BKM,2006, GRSTUX]
(A) 1 : 1 Ans.
(B) 1 :
2
(C*) 1 : 2
(D) 1 : 4
(C) For anti-clockwise motion, speed at the highest point should be
gR . Conserving energy at (1) & (2) :
(C)okekorZfn'kk esaxfrdsfy,] mPprefcUnq ijpky gR gksuhpkfg,A (1)rFkk(2)dse/;ÅtkZlaj{k.kls:
1 R 1 mv a2 = mg m(gR) 2 2 2
CIRCULAR MOTION
v a2 = gR + gR = 2gR
va =
2gR
For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not become loose any where until it reaches the peg B. nf{k.kkorZxfrdsfy,ckWcdsiklçkjEHkesadelsdebrukosxgksukpkfg,ftllsoks
tk;sA At the initial position :
T + mgcos600 =
çkjfEHkdfLFkfresa
mv c2 R
;
VC being the initial speed in clockwise direction. VCdhçkjfEHkdpkynf{k.korZfn'kkesagSA
138.
For
VC min : Put T = 0 ; VC min : ds fy, T = 0 j[kh gSA
VC =
gR 2
gR 2 = 1 2 2gR
VC/Va =
VC : V a = 1 : 2
Ans.
A ball attached with massless rope of the length swings in vertical circle as shown in figure. The total acceleration of the ball is a ˆi bˆj m/sec2 when it is at angle = 37º; (where a and b are positive constant). Find the magnitude of centripetal acceleration of the ball at the instant shown. The axis system is shown in figure.
,dxsandksyEckbZdhnzO;ekughujLlhlstksM+djÅ/oZo`Ùkesafp=kkuqlkj?kqek;ktkrkgSAtc;gÅ/oZls=37º dk dks.kcukjghgSrcbldkdqyRoj.k a ˆi bˆj m/sec2gS(tgk¡arFkkb/kukRedfu;rkadgS)bl{k.kijvfHkdsUnzh;Roj.k dkifjek.kcrkb,Av{kfudk;fp=kesan'kkZ;kx;kgSA [Made BKM - 2005] [M.Bank(0708)_CM_3.2]
Sol.
Normal or centripetal acceleration of the ball is component of net acceleration along PO.
Ans.
ac = a cos 53° + b cos 37° = 3a 4b 5 5
3a 4b + m/sec2 5 5
CIRCULAR MOTION
139.
A particle tied to a string of length l is given a velocity at lowest point which is insufficient to complete the circular path in the vertical plane. The other end of the string is fixed. The radius of curvature of the path just after the string slacks is: [M_Bank(07-08)_C.M._7.13] ,dyEckbZdhjLlhls,dd.kcka/kktkrkgSrFkkbldksfuEurefcUnqlsdqNosxnsdjNksM+rstksfdÅ/okZ/kjo`Ùkesa
iw.kZpØdsfy;svi;kZIrgSAjLlhdknwljkfljkfLFkjgSAjLlhds
Sol.
(A*) (B) 2 (C) /2 As the particle is in circular path upto that point, radius of curvature is .
(D) 3
Wheel A of radius rA = 10cm is coupled by a belt C to another wheel of radius rB = 25 cm as in the figure. The wheels are free to rotate and the belt does not slip. At time t = 0 wheel A increases it’s angular speed from rest at a uniform rate of /2 rad/sec2. Find the time in which wheel B attains a speed of 100 rpm. [ Hint: vA = vB] rA=10cmf=kT;kokysifg,dksnwljsrB=25cmf=kT;kokysifg;slscsYVC}kjkfp=kkuqlkjtksM+ktkrkgSAifg;s?kweus ds fy, LorU=k gS vkSj csYV fQlyrhugha gSAle;t=0ijAifg;kviuh dks.kh; pky fojkekoLFkkls ,dlekunj/ 2 rad/sec2 ls c<+kuk 'kq: djrk gSA og le; crkb;s tc B ifg;k 100 rpm dh pky çkIr dj ysrk gSA [ Hint: vA = vB] M.Bank_CM_1.17 [S-(06-07)_XIII (GRSTU)_DPP-74(a)_8]
[ Ans: 50/3 sec. ] For no slipping condition
rA A = rB B
B =
10 rA A = × rad/s2 25 5 rB
10 2 100 = rad/s 3 60 B = BO + Bt
B =
10 50 =0+ t t = sec 3 5 3
141.
A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration
of a particle P (shown in figure) on the ring is (6 i 8 j ) m/s2. At that instant angular acceleration of the ring is _______ & the angular velocity is _______. Radius of the ring is 2m. M.Bank_CM._1.16
fp=kkuqlkj ,doy; Zv{kds lkis{k?kw.kZu djrhgSA ?kw.kZuxfr dkryxygSAfdlh {k.koy;ijfLFkr ,dd.k P
( fp=kkuqlkj)dkRoj.k (6 i 8 j )m/s2 gSA bl{k.koy;dk dks.kh;Roj.k _________gSrFkk dks.kh; osx________
gSA¼oy;dhf=kT;k2ehgSa½
2 Ans. –3rad/sec kˆ , – 2 rad/sec kˆ .
142.
A particle is revolving in a circle with increasing its speed uniformly. Which of the following is constant? M.Bank_CM_1.45
,dd.kdhpkydks,dlekunjlsc<+krsgq,,do`Ùkh;iFkij?kqek;ktkrkgSArksfuEuesalsdkSulhjkf'kfu;rgksxh (A) centripetal acceleration (vfHkdsUnzh; Roj.k) (B) tangential acceleration (Li'kZjs[kh; Roj.k) (C*) angular acceleration (dks.kh; Roj.k) (D)noneof these(buesals dksbZugha)
CIRCULAR MOTION
respectively.
M.Bank_CM_1.58
,dd.k ,do`Ùkkdkj iFkesa fu;rpkyVlsxfrdjjgkgSA;fn ,dks.kh;foLFkkiugSrks=0, lsizkjEHkgksdj d.k dkjs[kh; laosx vf/kdre oU;wure gksxktcfddkeku Øe'k%gSA (A) 45º & 90º 144.
(B) 90º & 180º
(C*) 180º & 360º(D) 90º & 270º
Three particle A, B & C move in a circle of radius r =
1 m, in anticlockwise direction with speeds 1 m/s, 2.5
m/s and 2 m/s respectively. The initial positions of A, B and C are as shown in figure. The ratio of distance travelled by B and C by the instant A, B and C meet for the first time is 1
rhu d.k A,BrFkkCf=kT;k r= m,ds o`Ùk ij okeko`r fn'kk esa Øe'k% 1m/s,2.5m/srFkk2m/s ls xfreku gSA A, BrFkkCdhçkjfEHkdfLFkfr;kafp=kesan'kkZ;hxbZgSABrFkkC}kjkr;dhxbZnwfj;ksdkvuqikrml{k.kgS]tcA,B rFkk Cigyhckj feyrsgS &
M.Bank_C.M._1.9
(A) 3 : 2 145.
(B*) 5 : 4
(C) 3 : 5
(D) 3 : 7
A particle is revolving in a circle of radius R with initial speed v. It starts retarding with constant retardation v2 8 R . The number of revolutions it makes in time is: 4 R v
M.Bank_C.M._1.46
v2 ,dd.kRf=kT;kdso`ÙkesaizkjfEHkdpkyvlsxfr'khygSA;gfu;renau4 R dslkFkeafnrgksukizkjEHkdjrkgSA;g 8 R le;esa fdruspDdj iwjsdjsxk & v
(A) 3
(B) 4
(C*) 2
(D)noneof thesemijksDr esa ls dksbZ ugha
Passage # 1 (146 - ) [Made J.K.Sir 2006, GRSTUX] BEK_20.8.06_comp1 (Read the following passage and answer the questions numbered 31 to 35. They have only one correct option) Circus Act A new circus act was developed by Gemini circus. Riya (a circus girl) swings from a trapeze, projects herself at an angle of 53º as shown and supposed to be caught by Ravi, whose hands are 3.2m above and 4.8 m horizontally from her launch point. Once Riya is projected, she moves freely under gravity. She requires certain minimum velocity ‘v 0’ to reach Ravi. This velocity may be evaluated by using equation of motion in 2-D. Once Riya reaches to Ravi, she is caught by Ravi. Now both move with same speed upward and both swing around suspension point ‘O’.
vuqPNsn #1 (fuEu vuqPNsn dks /;kuls i<+srFkk iz'u la[;k 31ls 35rdmRrj nsaA buds dsoy ,d lR;fodYi gS) ldZl&[ksy tsfeuhldZlesau;kldZl&[ksyfodflrfd;kx;kAldZlyM+dhfj;keqäjLlhls?kweldrhgSAog53ºdks.kijiz{ksfir gksrhgS]rFkkjfo}kjkidM+htkrhgSAjfodsgkFkiz{ksi.kfcUnqls3.2eh-Å¡pkbZrFkk4.8eh-{kSfrtnwjhijgSaA ,dckjfj;kiz{ksfirgksrhgS]ogxq:RoizHkkoesaeqä:ilsfxjrhgSAmls]jfordigq¡pusdsfy,U;wureosx‘v0’ dhvko';drk gksrhgSA blosx dhx.kuk xfrds f}&foeh;lehdj.kksa }kjkdh tkldrh gSA,d ckjtc fj;k]jfo rd igq¡prhgS]rksogmlsidM+rkgSAvcnksuksalekupkylsÅijdhvksjpyrsgSarFkknksuksafuyEcufcUnq‘O’dsifjr%?kwers
CIRCULAR MOTION
g S a A
In order to safe guard their lives, a safety net is provided 4m below the launching point of Riya. For all the following calculations take g = 10 m/s2.
nksuksadhftUnxhlqjf{krdjusdsfy,],dlqj{kktky]iz{ksfirfcUnqls4eh-uhpsyxkrsgSAfuEulHkhx.kukdsfy,g =10eh-@ls-2 ysaA 146.
In order to just reach Ravi, the initial speed of Riya should be equal to :
jfordigq¡pusdsfy,]fj;kdkvko';di;kZIrosxgksukpkfg,& Sol.
(A) 45 km/hr (D) 24 km/hr Trajectory equation : y = x tan –
3.2 = 4.8 ×
(C*) 36 km/hr
iFkdslehdj.kls
g x2 2 v 02 cos2
10 4.8 4.8 25 4 – 2 v 20 9 3
v 02 =
10 4.8 4.8 25 2 9 3.2
v 02 = 4 × 25, 147.
(B) 30 km/hr
v 0 = 10m/sec.
= 36 km/h.
Ans.
In the above problem, at the instant when Riya reaches to Ravi, the magnitude and direction of her velocity is :
mijksäiz'uea]fj;kdsjfordigq¡pusds{k.k]fj;kdsosxdkifjek.kofn'kkgS& (A) 21.6 km/hr, at 37° upward from horizontal. 21.6 km/hr] {kSfrt ls 37ºds dks.kij Åijdh vksjA (B*) 21.6 km/hr, in horizontal direction. 21.6km/hr,{kSfrtfn'kkesa (C) 16.2 km/hr, at 37° upward from horizontal. 16.2 km/hr] {kSfrt ls 37ºds dks.kij Åijdh vksjA (D) 16.2 km/hr, in horizontal direction. 16.2km/hr{kSfrtfn'kkesa Sol.
x = v 0 cos × t 4.8 = 10 ×
3 ×t 5 t = 0.8 sec.
v y = v 0 sin – gt = 10 ×
4 – 10 × 0.8 = 8 – 8 = 0 5
CIRCULAR MOTION
as v y = 0, then motion is in horizontal direction and velocity is v 0 cos = 10 × =6× 148.
18 5
= 21.6 km/hr
3 = 6 m/sec. 5
Hence (B).
In the above problem, if mass of Ravi is ‘2m’ and mass of Riya is ‘m’, then when riya is caught by Ravi both move together with the speed equal to :
mijksäiz'uesa];fnjfodknzO;eku‘2m’rFkkfj;kdknzO;eku‘m’gSAtcjfo]fj;kdksidM+rkgS]rcnksuksalkFk&lkFk fuEuesadkSulhpkylspyrsgSa\ Sol.
(A) 9 km/hr (m + 2m) v = mv 0 cos 3mv = m × 10 ×
(B) 6 km/hr
(C*) 7.2 km/hr
(D) 4.8 km/hr
3 5
v = 2 m/sec 149.
If both move together further and both of them are considered as one mass system of 3m. Effective distance of this one mass system from suspension point is assumed to be 1 meter, then angle through which they deflect will be :
Sol.
;fnnksuksadks3mnzO;ekudk,dfudk;ekusa]rFkknksuksavkxspyrsgSaA;fnblnzO;ekudhfuyEcufcUnqlsizHkkohnwjh 1eh-ekusa]rcdks.kftllslsfo{ksfirgksaxs& (A) 30° (B*) 37° (C) 53° (D) 60° v 2 = 2gl [1 – cos] 2 × 2 = 2 × 10 × 1 [1 – cos] 1 – cos =
150.
1 4 , cos = 5 5 = 37°
In their debut performance, once Ravi misses the Riya completely as she flies past. The horizontal distance through which Riya moves from the initial launch point before landing in the safety net 4m below her initial launch point is :
mudsigysizn'kZuesa],dckjjfo]fj;kdksidM+usesapwdtkrkgSrFkkogvkxspyhtkrhgSrksfj;k}kjkizkjfEHkdiz{ksi.k fcUnqls4muhpslqj{kktkyesafxjusrdpyhxbZ{kSfrtnwjhgS& (A) 10 m Sol.
(B) 9 m Given y = 4 m
fn;kgS – y = x tan –
(C*) 12 m
(D) 9.6 m
g x2 2 v 02 cos2
4x 10 x 2 25 – 3 2 10 10 9 – 36y = 48 x – 5 x2 5x2 – 48x – 36 y = 0 5x2 – 48x – 144 = 0
–y=
5x2 – 60x +12x – 144 = 0 5x (x – 12) + 12 (x – 12) = 0 x = 12 m 151.
Ans.
The velocity and acceleration vectors of a particle undergoing circular motion are v = 2ˆi m/s and a = 2ˆi + 4ˆj m/s2 respectively at an instant of time. The radius of the circle is fdlh o`r esa pDdj dkV jgs d.k ds fdlh le; osx lfn'k v = 2ˆi m/srFkk Roj.k lfn'k a = 2ˆi + 4ˆj m/s2 gSA bl
CIRCULAR MOTION
o`rdhf=kT;kgS& (A*) 1m Sol.
(B) 2m
(C) 3m (D) 4m [Made MPS-2005] [M.Bank_CM_2.14] It can be observed that component of acceleration perpendicular to velocity is a = 4 m/s2 c
152.
radius =
v2 (2)2 = = 1 metre. ac 4
A car is moving on circular path of radius 100m such that its speed is increasing at the rate of 5m/s2. At t = 0 it starts from rest. The radial acceleration of car at the instant it makes one complete round trip, will be _______. M.Bank_CM_1.27 ,d dkj f=kT;k 100 ehVj ds iFk ij bl izdkj xfr djrh gS fd bldh pky 5 m/s2 dh nj ls c<+ jgh gSA t = 0 le;
ij;gfojkels 'kq:gksrhgSAtcdkj,d pDdjiwjk djrhgSrksdkj dkf=kT;h; Roj.kgksxk& [Ans: 20 m/s2] 153.
A circular road of radius R is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circular road, the friction co-efficient between the tyre & road is negligible: [6.6_Circular Motion] R f=kT;k dh ,d o`Ùkkdkj lM+d dks pky V= 40km / hrds fy;s cafdr fd;k tkrk gSA m nzO;eku dh ,d dkj bl
o`ÙkkdkjlM+dijtkrhgSAVk;jolM+ddse/;?k"kZ.k xq.kkadux.;gSA (A) the car cannot make a turn without skidding (B) if the car runs at a speed less than 40 km/hr, it will slip up the slope (C) if the car runs at the correct speed of 40 km/hr, the force by the road on the car is equal to mv 2/r (D*) if the car runs at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv 2/r (A)dkjfcukfQlysaugheqM+ldrhgSA (B) ;fn dkj 40 km / hr ls de pky ls pyrh gS rks ;g <+ky ij Åij dh vkjs fQlysxh A (C) ;fn dkj lgh pky 40 km / hr ls pyrh gS rks lM+d }kjk dkj ij yxk;k x;k cy mv2/r gksxk A (D*) ;fn dkj lgh pky 40 km/hr, ls pyrh gS rks lM+d }kjk dkj ij yx;k x;k cy mgrFkk mv2/nksuksa ls cM+k
gksxkA 154.
A ball tied to the end of a string swings in a vertical circle under the influence of gravity
fdlhMksjhds,dfljslsca/kh,dxsanxq:Rodsv/khum/okZ/kjo`ÙkesaxfrdjrhgSA (A) when the string makes an angle 90º with the vertical , the tangential acceleration is zero & radial acceleration is somewhere between maximum and minimum (B*) when the string makes an angle 90º with the vertical, the tangential acceleration is maximum & radial acceleration is somewhere between maximum and minimum (C) at no place in the circular motion, tangential acceleration is equal to radial acceleration (D*) throughout the path whenever radial acceleration has its extreme value, the tangential acceleration is zero. [3.56_Circular Motion] (A)tcMksjhm/okZ/kjdslkFk90ºdksdks.k cukrhgSrcLi'khsZ Roj.k 'kwU;rFkkf=kT; Roj.kvf/kdre oU;qureds
chpgSA (B*)TkcMksjhm/okZ/kjdslkFk90ºdkdks.kcukrhgSrcLi'khZRoj.kvf/kdrerFkkf=kT;Roj.kvf/kdreoU;qure ds chpgSA (C)lEiw.kZ &o`Ùkh;xfresadghijHkhLi'khZRoj.k]f=kT; Roj.kdsacjkcjughagskrkgSA (D*)LkEiw.kZZiFk esatcdHkhf=kT;Roj.k viuspjeekuijgksrkgSrksLi'khZ Roj.k 'kwU;gskrkgSA 155.
A ball suspended by a thread swings in a vertical plane so that its acceleration magnitudes in the extreme and lowest positions are equal. The threads deflection angle in the extreme position is___________. [3.58_Circular Motion]
,dxsan,d/kkxs}kjkÅ/oZryesablçdkjxfrdjrhgSfdxfrdsvfUre,oafuEurefLFkfr;ksaijbldkRoj.kleku
CIRCULAR MOTION
gSAvfUre(external)fLFkfrij/kkxsdkfo{ksidks.kgksxk___________
156.
4 [ Ans: 1.86 Ir sin-1 ~ 53º] 5 A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is x2 = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is : ,dm nzO;ekudkchM+ijoyf;drkjijgSrFkkbldhv{kÅ/okZ/kjgSrFkk'kh"kZ(vertex)ewyfcUnqijfp=kkuqlkjgSA,oe~ bldh lehdj.k x2 = 4ay}kjknh tkrhgSA rkjdk ÝsefLFkj(fixed)gSrFkk bl ij chM+ fcuk ?k"kZ.kds fQyldrk gSA chM+dksfcUnq y=4alsfojkelsNksM+ktkrkgSAblchM+dkLi'kZ&js[kh;Roj.kD;kgksxktc;gfcUnqy=aijigq¡prk gS: [Q. 1.8_CM] BM_CM_161
[Q.161/RK_BM/Circular Motion] [Made MPS, 2005]
g 3g (B) 2 2 x2 = 4ay Differentiating w.r.t. y, we get
(A) Sol.
(C*)
g 2
(D)
g 5
dy x = dx 2a dy =1 dx hence = 45° the component of weight along tangential direction is mg sin .
At (2a, a),
hence tangential acceleration is g sin = 157.
g 2
The square of the angular velocity of a certain wheel increases linearly with the angular displacement during 100 rev of the wheel's motion as shown. Compute the time t required for the increase. ,difg;sdhxfrdsnkSjkumldsdks.kh;osx dkoxZ]dks.kh;foLFkkiudslkFkjs[kh;:ilsifg;sds 100pDdj ds nkSjku fp=kkuqlkj cnyrk gSA bldh o`f} ds fy, vko';d le; t dh x.kuk dhft, &
[Q. 1.25_CM]
[ Ans : 158.
40 sec. ] 7
A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating and is [Q. 3.10_CM]
CIRCULAR MOTION
,dd.ktksfcUnqAij,dfLFkjfpdusrf=kT;kdsv)Zxksysijfp=kkuqlkjfojkekoLFkkesagSAfcUnqBij;gv)Zxksysls laidZNksM+nsrkgSACv)ZxksysdkdsUnzgSrFkkdkslEc)djusokyhlehdj.kgksxh-
[Made 2005, MPS]
(A) 3 sin = 2 cos (B) 2 sin = 3 cos (C*) 3 sin = 2 cos (D) 2 sin = 3 cos Sol.(C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv 2 = mg sin .......... (1) r Applying conservation of energy as the block moves from A to B..
1 mv 2 = mg (r cos – r sin ) 2 Solving 1 and 2 we get 3 sin = 2 cos
159.
A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle is rotated about diameter AB as axis with a speed as shown in figure. The bead P is at rest w.r.t. the circular ring in the position shown. Then 2 is equal to: [Q. 2.17_CM] ,dfpdusrkjdksaf=kT;kdsm/okZ/kjo`ÙkesaeksM+rsgSA,deudkP,rkjijfQlyldrkgSAo`Ùkfp=kkuqlkjO;klAB dslkis{kpkyls?kw.kZudjrkgSAfp=kesfn[kkbZfLFkfresaeudkPo`Ùkkdkjoy;dslkis{kfLFkjgSArks2cjkcjgS&
(A)
160.
.......... (2)
2g a
(B*)
2g a 3
(C)
g 3 a
In the above question the centripetal acceleration of the particle is:
(D)
2a g 3
[Q. 2.18_CM]
mijksDriz'uesad.kijvfHkdsUnzh;cygS (A) 161.
5 3 2
(B) 5 3
(C)
5 2 3
(D*)noneof thesebuesalsdksbZugha
A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating and is [Q. 3.10_CM] ,dd.ktksfcUnqAij,dfLFkjfpdusrf=kT;kdsv)Zxksysijfp=kkuqlkjfojkekoLFkkesagSAfcUnqBij;gv)Zxksysls laidZNksM+nsrkgSACv)ZxksysdkdsUnzgSrFkkdkslEc)djusokyhlehdj.kgksxh-
[Made 2005, MPS]
(A) 3 sin = 2 cos (B) 2 sin = 3 cos (C*) 3 sin = 2 cos (D) 2 sin = 3 cos Sol.(C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface.
CIRCULAR MOTION
mv 2 = mg sin .......... (1) r Applying conservation of energy as the block moves from A to B..
1 mv 2 = mg (r cos – r sin ) 2 Solving 1 and 2 we get 3 sin = 2 cos
162.
A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle is rotated about diameter AB as axis with a speed as shown in figure. The bead P is at rest w.r.t. the circular ring in the position shown. Then 2 is equal to: [Q. 2.17_CM] ,dfpdusrkjdksaf=kT;kdsm/okZ/kjo`ÙkesaeksM+rsgSA,deudkP,rkjijfQlyldrkgSAo`Ùkfp=kkuqlkjO;klAB dslkis{kpkyls?kw.kZudjrkgSAfp=kesfn[kkbZfLFkfresaeudkPo`Ùkkdkjoy;dslkis{kfLFkjgSArks2cjkcjgS&
(A)
163.
.......... (2)
2g a
(B*)
2g a 3
(C)
g 3 a
In the above question the centripetal acceleration of the particle is:
(D)
2a g 3
[Q. 2.18_CM]
mijksDriz'uesad.kijvfHkdsUnzh;cygS (A) 164.
(B) 5 3
(C)
5 2 3
(D*)noneof thesebuesalsdksbZugha
A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end of A of the rod. The rod is set in angular motion about A with constant angular acceleration If the coefficient of friction between the rod and the bead is and gravity is neglected, then the time after which the bead starts slipping is: ,dyEcs{kSfrtNM+ij,deudkgStksbldhyEckbZdsvuqfn'kxfrdjrkgSvkSjçkjEHkesafcUnqAlsLnwjhijj[kk gSANM+dksfcUnqAdslkis{kfu;rdks.kh;Roj.klsxfrdjkbZtkrhgSAvxjeudsrFkkNM+dschp?k"kZ.kgSrFkk xq:Ro dks ux.; ekuk tk, rks og le; ftlds ckn fQlyu :d tk,xk : M.Bank_CM_1.33 (A*)
165.
5 3 2
(B)
(C)
1
(D) infinitesimal cgqr NksVk
A car driver going at some speed suddenly finds a wide wall at a distance r. To avoid hitting the wall he should M.Bank_CM_1.56
fdlhpkylspyrkgqvk,ddkjpkydrnwjhij,dpkSM+hnhokjns[krkgSA nhokjijVdjkuslscpusdsfy;smls (A*) apply the brakes (B) should turn the car in a circle of radius r. (C) apply the brakes and also turn the car in a circle of radius r. (D) jump on the back seat. (A*)cszdyxkkus pkfg;sa (B) dkjdksrf=kT;k ds,do`ÙkesaeksM+ukpkfg;sA (C)csyyxkuspkfg;svkSjdkjdksrf=kT;k dso`ÙkesaHkheksM+ukpkfg;sA (D) dkj dh fiNyh lhV ij dwn tkuk pkfg;s A
CIRCULAR MOTION
166.
A particle of mass m describes a circle of radius r. The centripetal acceleration of the particle is 4/r2. What will be the momentum of the particle ? M.Bank_CM_1.51 m nzO;ekudk,dd.krf=kT;kdso`resaxfr'khygSAd.kdkvfHkdsUnzh;Roj.k4/r2gSaAd.kdklaosxD;kgksxkA (A) 2
167.
m r
(B*) 2
m r
(C) 4
m r
(D) none
A ring of mass 2 kg and of radius 0.25 m is making 300 rpm about an axis through its centre perpendicular to its plane. The tension (in newton's) developed in the ring is (take 2 = 10) M.Bank_CM_1.52
,d 2kg nzO;eku o 0.25 m f=kT;k dh oy;] blds ry ds yEcor o dsUnz ls xqtjus okyh v{k ds lkis{k 300rpm ls ?kw.kZudjrhgSAoy;esamRiUurukogksxk(U;wVuesa2=10ekusa) (A) 50
(B) 100
(C) 175
(D*) 250
168.
In the figure shown a lift goes downwards with a constant retardation. An observer in the lift observers a conical pendulum in the lift, revolving in a horizontal circle with time period 2 seconds. The distance between the centre of the circle and the point of suspension is 2.0 m. Find the retardation of the lift in m/s2. Use 2 = 10 and g = 10 m/s2 [Made 2005 RKV] M.Bank_CM_2.1 fn,x,fp=kesafy¶Vfu;reanulsuhpsdhvkSjxfrdjjghgSA,dizs{kdtksfy¶VdsvUnjgS]2lSd.MdsvkorZ&dky ls{kSfrto`ÙkesapDdjdkVjgkgSAo`rdsdsUnz,oe~yVdu fcUnqdschpnwjh2.0mgSAfy¶VdkeanuKkrdhft,A mi;ksx esa ys2 = 10 vkSjg = 10 m/s2 [Made 2005 RKV] M.Bank_CM_2.1
Sol.
T = 2
cos geff . = 2
h geff .
geff. = g + a ; put T = 2 a = 10 m/s2. Ans.
Retardation = 10 m/s2
Ans. 10 169.
Sol.
170.
A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are a = 2 ˆi + 3 ˆj m/s2 and P = 6 ˆi – 4 ˆj kgm/s. Then the motion of the particle is ,d d.k o`Ùkh; iFk ij xfr dj jgk gSA fdlh le; Roj.k rFkk laosx lfn'ka = 2 ˆi + 3 ˆj m/s2 rFkk P = 6 ˆi – 4 ˆj kgm/ sgSaAd.kdhxfrgS(A) uniform circular motion (B) circular motion with tangential acceleration (C) circular motion with tangential retardation (D*) we cannot say anything from a and P only.. (A) leku o`rh; xfr (B)Li'kZjs[kh;Roj. klso`Ùkh;xfr (C)Li'kZjs[kh;eanulso`Ùkh;xfr (D*) dsoy a rFkk P ls ge dqN ugha dg ldrs [Made 2005, MPS] [1.6_Circular Motion] The nature of the motion can be determined only if we know velocity and acceleration as function of time. Here acceleration at an instant is given and not known at other times so D. Wheel A of radius rA = 10cm is coupled by a belt C to another wheel of radius rB = 25 cm as in the figure. The wheels are free to rotate and the belt does not slip. At time t = 0 wheel A increases it’s angular speed from rest at a uniform rate of /2 rad/sec2. Find the time in which wheel B attains a speed of 100 rpm. [ Hint: vA = vB] rA=100cmf=kT;kokysifg,dksnwljsrB=25cmf=kT;kokysifg;slscsYVC}kjkfp=kkuqlkjtksM+ktkrkgSAifg;s?kweus ds fy, LorU=k gS vkSj csYV fQlyrh ugha gSA le; t= 0ij ifg;k viuh dks.kh; pky t= 0ls /2rad/sec2 c<+uk 'kq: djrk gSA og le; crkb;s tc ifg;k 100 rpm dh pky çkIr dj ysrk gSA [ Hint: vA = vB]
CIRCULAR MOTION
M.Bank_CM_1.17
[ Ans: 50/3 sec. ] 171.
A car initially traveling eastwards turns north by traveling in a quarter circular path of radius R metres at uniform speed as shown in figure. The car completes the turn in T second. ,ddkjfp=kesan'kkZ,vuqlkjiwoZdhvksjxfrdjjghgS]fp=kesan'kkZ,vuqlkjRf=kT;kdspkSFkkbZo`Ùkh;iFkijxfr djrhgqvk mÙkjdh vksjeqM+ tkrhgSA dkjpDdjiwjkdjus esa TlSd.Mdk le;ysrhgSA (a) What is the acceleration of the car when it is at B located at an angle of 37. Express your answers in terms of unit vectors ˆi and ˆj
dkjdkRoj.kD;kgksxktc;gfp=kkuqlkj37ºdks.kijfLFkrgSAvkidkmÙkj,dkadlfn'k ˆi rFkkˆj dsinksa esanhft,A M.Bank_Circu_Motion._1.3 (b)
Sol.
The magnitude of car's average acceleration during T second period. T lSd.Mle;dsnkSjkudkjds vkSlrRoj.kdkifjek.k gksxk?
Speed of car is v =
R m/s 2T
[Made MPS - 2005]
.....
2 R v2 (a) The acceleration of car is = at B and is directed from B to O. 20 T 2 R Acceleration vector of car at B is 2 R v2 ˆ ˆ = ( – sin 37° + cos 37° ) = (– 3 ˆi + 4 ˆj ) m/s2 j i a 20 T 2 R (b) The magnitude of average acceleration of car is in time T is R v C VB 2v = = m/s2 2 T2 T T
Comprehension (172 - 176) M.Bank_CM_Comp. # 2 (Read the following passage and answer the questions numbered 11 to 15. They have only one correct option) In a certain experiment to measure the ratio of charge and mass of elementary charged particles, a surprising result was obtained in which two particles moved in such a way that the distance between them remained constant always. It was also noticed that, this two particle system was isolated from all other particles and no force was acting on this system except the force between these two masses. After careful observation followed by intensive calculation it was deduced that velocity of these two particles was always opposite in direction and magnitude of velocity was 103 m/s and 2 × 103 m/s for first and second particle respectively and masses of these particles were 2 × 10–30 kg and 10–30 kg respectively. Distance between them came out to be 12 Å . (1 Å = 10–10 m) [Made RS 2007, A1-A8]
vuqPNsn (fuEu vuqPNsndks /;kuls i<+srFkk iz'ula[;k 11ls 15rdmRrj nsaAbuds dsoy,d lR;fodYi gSa)
ewyvkosf'krd.kksadsvkos'k,oanzO;ekudhfu"ifRrekiusdsfdlhiz;ksxesa,dvk'p;Ztfurifj.kkeizkIrgqvk]ftlesa nksd.kblrjgxfrekugSa]ftllsmudschpdhnwjhges'kkfu;rjgrhgSAiz;ksxesa;gHkhik;kx;kfd;gnksd.k fudk;vU;lHkhd.kksalsvyxFkkrFkkblfudk;ijnksnzO;ekuksadschpyxusokyscydsvfrfjävU;dksbZcydk;Zjr ughaFkkA/;kuiwoZijh{k.krFkkl?kux.kukdsi'pkr~nksuksad.kksadsosxges'kkfoijhrfn'kkesaik;sx;srFkkftudkifjek.k
CIRCULAR MOTION
Øe'k% izFked.kdsfy,103 eh-@ls- rFkkf}rh;d.k dsfy,2× 103
[email protected];eku Øe'k%2×10– 30 fdxzk-rFkk 10–30 fdxzk- FksA muds chp dh nwjh 12ÅKkr dh xbZA(1Å= 10–10 eh-) [Made RS 2007, A1-A8] 172.
Acceleration of the first particle was
izFked.kdkRoj.kFkk& (A)zero'kwU; Sol.
(B) 4 × 1016 m/s2
(C) 2 × 1016 m/s2
(D*) 2.5 × 1015 m/s2
(D) The two particle move in different circles. The mutual interaction force provides the required centripetal force to the particle. As magnitude of the interaction force is same F12 =
m1v12 m2 v 22 and F21 = r1 r2
F1 = F2
m1v12 m2 v 22 = r1 r2
putting values, we get r2 = 2 r1 Also r1 + r2 = 12 × 10–12 m (given) r1 = 4 × 10–12 m & r2 = 8 × 10–12 m Acceleration of first particle = 173.
(10 3 m / s)2 v 12 = = 2.5 × 1015 m/s2 r1 ( 4 10 12 m)
Acceleration of second particle was
f}rh;d.kdkRoj.kFkk& Sol.
(A*) 5 × 1015 m/s2 (A)
(B) 4 × 1016 m/s2
Acceleration of second particle is = 174.
(C) 2 × 1016 m/s2
(D)zero'kwU;
v 22 (2 10 3 )2 = = 5 × 1015 m/s2 r2 (8 10 12 )
If the first particle is stopped for a moment and then released. The velocity of centre of mass of the system just after the release will be :
;fnizFked.kdks{k.kHkjdsfy,jksdsarFkkeqädjnsaArkseqäNksM+usdsrqjUrckn]fudk;dsnzO;ekudsUnzdkosxgksxk& (A) Sol.
1 × 10–30 m/s 3
Sol.
1 × 103 m/s 3
(C*)
2 × 103 m/s 3
(D)noneof thesebuesalsdksbZugha (C) Just after release : VCM =
175.
(B)
m1v1 m 2 v 2 ( 2 10 30 )(0) (10 30 )(2 10 3 ) 2 = = × 103 m/s 30 m1 m 2 3 3 10
Path of the two particles was (A) Intersecting straight lines (C*) Circular
(B) Parabolic (D) Straight line w.r.t. each other
nksuksad.kksadkiFkFkk& (A)ijLijdkVrhgqbZnksljyjs[kk,saA (C*)o`ÙkkdkjA
(B) ijoy;A (D),dnwljsdslkis{kljyjs[kk
(C) Since the distance between them always remains constant but
CIRCULAR MOTION
move with different velocities. Therefore they must move in different circles with common centre as shown in the figure.
176.
Angular velocity of the first particle was :
izFked.kdkdks.kh;osxFkk& Sol.
(A*) 2.5 × 1012 rad/s (A)
(B) 4 × 1012 rad/s
Angular velocity of the first particle was 1 = 177.
(C) 4 × 1013 rad/s
(D)zero'kwU;
v1 (10 3 m / s) = r1 (4 10 12 m)
1 = 2.5 × 1012 rad/s
Two blocks A and B each of same mass are attached by a thin inextensible string through an ideal pulley. Initially block B is held in position as shown in figure. Now the block B is released. Bolck A will slide to right and hit the pulley in time tA. Block B will swing and hit the surface in time tB. Assume the surface as frictionless. nksxqVdsArFkkBftudknzO;ekulekugS]fp=kkuqlkjjLlhls,dvkn'kZiqyh}kjktqM+sgq,gSAçkjEHkesaxqVdsBdks fp=kkuqlkjj[kktkrkgSAvcxqVdsB dksNksM+ktkrkgSAxqVdkAnka;hvksjfQlyrkgSvkSjiqyhdkstAle;esatkdjVdjkrk gSA xqVdk Blrg ij ] le; tB esa tkdj Vdjkrk gSA ;g ekfu, fd lrg ?k"kZ.kghu gS– M.Bank_Circular Motion_3.34
(A) tA = tB (B*) tA < tB (C) tA > tB (D) data are not sufficient to get relationship between tA and tB. vk¡dM+stArFkktB dse/;lEcU/kfudkyusdsfy,vi;kZIrgSA 178.
Two particles A & B separated by a distance 2 R are moving counter clockwise along the same circular path of radius R each with uniform speed v. At time t = 0, A is given a tangential acceleration of magnitude a =
72 v 2 . Find : 25 R
nksd.kArFkkBftudschpnwjh2RgS],dlekuo`Ùkh;iFkftudhf=kT;kRgS]?kM+hdsxfrdhfn'kkdsfoijhrfn'kk esa xfrdjjgsgSAle;t= 0ij Adks Li'kZjs[kh;Roj.kftldkifjek.ka= (i) (iii) (i) (iii)
the time lapse for the two bodies to collide angular velocity of A
le;tcnksuksoLrq,aVdjkrhgSA Adk dks.kh;osxA
[ Ans: (i) 179.
11 5R sec (ii) 6 6v
(iv)
(iii)
72 v 2 gS]fn;k tkrkgS@crkb, 25 R
M.Bank_Circular Motion_1.18 (ii) the angle covered by A (iv) radial acceleration of A. (ii) blle;esaA}kjkcuk;kx;kdks.kA Adkdks.kh;Roj.k
17 v 289 v 2 (iv) ] 5R 25 R
A force of constant magnitude F acts on a particle moving in a plane such that it is perpendicular to the velocity v ( | v | = v) of the body, and the force is always directed towards a fixed point. Then the angle turned by the velocity vector of the particle as it covers a distance S is :(take mass of the particle as m) [Made PKS, 2005] M.Bank_CM._1.7 ,d fu;r ifjek.k dk cy F,lery esa xfr djrs d.k ij bl izdkj yxrk gS fd ;g oLrq ds osx v (| v | = v)ds yEcor~
CIRCULAR MOTION
yxrkgSvkSjcyges'kkfuf'prfcUnqdhvksjyxrkgSrksosxlfn'k}kjk?kwekx;kdks.kgksxktcrdd.kSnwjhr; djrk gS (d.kdk nzO;ekumysaA) (A) Sol.
FS 2
2FS
(C)
2
2mv mv Since F V , the particle will move along a circle.
FS2 mv
(D*)
FS mv 2
FS S mv 2 & = = R R mv 2 Two particles tied to different strings are whirled in a horizontal circle as shown in figure. The ratio of lengths of the strings so that they complete their circular path with equal time period is: [Made 2006, SNS, GRSTX] M.Bank_CM._1.63
F=
180.
(B)
jLlhlscU/ksnksd.kksadksfp=kkuqlkjo`ÙkkdkjiFkesa?kwek;ktkrkgSAjfLl;ksadhyEckbZdkvuqikrgSftllsosviuko`Ùkh; iFkleku le;dky esaiwjk djrsgS &
(A)
Sol.
181.
3 2
(B*)
since
T = 2
2 3
(C) 1
(D) None of these
L cos g
T1 = T2
L1 cos 2 L 2 cos 1
=
L1 cos1 = L2 cos2
cos 45 cos 30
L1 L2
2 3
A 10kg ball attached at the end of a rigid rod of length 1m rotates at constant speed in a horizontal circle of radius 0.5m and period 1.57 s as shown in the figure. The force exerted by the rod on the ball is M.Bank_CM._2.24
1myEchn`<+NM+ds,dfljsijca/kh10kgdhxsan 0.5mf=kT;kds{kSfrto`Ùkesfu;rosxls?kwerhgSA¼fp=kkuqlkj½ bldk vkorZdky 1.57sgSA NM+ }kjk xsan ij yxk;k x;k cy gksxk \
(A) 158 N 182.
(B*) 128 N
(C) 110 N
(D) 98 N (g = 10 ms-2)
A particle moving on the inside of a smooth sphere of radius r describing a horizontal circle at a distance r/ 2 below the centre of the sphere. What is its speed? M.Bank_CM._2.26 ,dd.krf=kT;kdsfpdusxksysesaxksysdsdsUnzlsr/2nwjhuhps{ksfrto`ÙkesaxfrdjjgkgSAbldhpkyD;kgksxh\ (A)
5 gr
(B)
4 gr 3
(C*)
3 gr 2
(D)
3 gr
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CIRCULAR MOTION
