This is the teaching module for circular motion. This module has to be followed in the class. VK Bansal
CIRCULAR MOTION (KINEMATICS) LECTURE In XIII 3½ Lectures in Bull's Eye 4+1 Lectures & ACME 4 Lectures including discussion. 1.
INTRODUCTION: Lets say you are watching F-1 rate Michael Schumacher is going on a curved path and you want to sheet him by your handycam. If you are given two positions to stand O1 & O2 which one will you choose?
Ans.
O2 why? 2 < 1 hence easy manouvering at camera you have to cover less angle in same time. Although you may not have much idea about circular motion but your decision was based on analysis of angular variables.You decided by thinking that you have to cover smaller angle in same time.(2 and 1 are angle substended by M.S. while going from P1 to P2 on O2 and O1)
2.
ANGULAR VARIABLES: Angular Displacement: Angle substended by a moving particle on a fixed point is called angular displacement about the fixed point. Thus in above discussion angular displacement about O1 is 1 & about O2 is 2. Few Facts: Dimensionless quantity Units – radian / Never Degree
Imp.
Angular displacement depends on reference frame same as linear displacement depends on refrence frame. Angular displacement will be different if we observer the car from another car. (but angular displacement is different for different observers in the same frame). (The linear displacement is same for two observers at different positions in same frame) e.g. O1 & O2 will observe same linear displacement but different angular displacement although both points are in the same ground frame.
(V.Imp) It may be a bit shocking for you but it is a vector quantity. Direction of angular displacement vector is decided by right hand rule.i.e. move your right hand fingers in sense of motion and direction of your thumb will be the direction of angular displacement.
It sounds quite confusing that direction of vector is nowhere near the actual motion. But if you pay attention to a few facts you will understand that perhaps this is the best way to represent angular displacement. If a vector represents angular displacement it holds three informations which can completely describe the angular displacement.(Take a pen show it to students tell them this is the angular displacement vector and then deduce these three facts giving them full idea of 3–D view) Page # 1
The unique plane perpendicular to the line represents the plane of motion of particle. Now place your right hand thumb along the vector and direction of your fingers will give you the sense of rotation. Length of the vector will represent the magnitude angular displacement. i.e. length of vector representing four complete rotations will be 8 and representation of one complete revolution will be by length of 2. Note: Above discussion was for increasing comfort level of students with angular variables tell them following as well. In JEE syllabus plane of circular motion is fixed so the direction of angular variables remains same (although sense may become +ve or –ve). Thus even though they are vectors but we need not give much thought to it as there directions will remain unchanged. Much like one-dimensional motion where variables are vectors but we do not use vector notation, we simply use signs +ve & –ve to represent sense. In short we will study angular kinematics of one-D. i.e. will never use vector notation but stick to sign +ve or –ve by defining our +ve sense. Angular velocity: Rate of angular displacement is called angular velocity. =
d dt
Unit rad s–1 / Dimension T–1 Relation between angular velocity & linear velocity. V cos P
Q a
P' V
O V sin
A particle P is moving with speed V along a curve & observer is located at O - is angle between line joining OP and velocity. Note that V sin (comp. of velocity perpendicular to OP) is the cause of angular displacement. i.e. if only V cos existed we need not turn our head to always look at particle. Hence PQ = (V sin ) t PQ = OP ()
Thus
t
=
V sin (OP )
Comp. of velocity perpendicular to line joining length of line joining
Asking Question In circular motion find angular velocity of particle moving with speed V wrt center
= Ex.
also, v r
V R
A projectile (u,) is launched from horizontal plane, find angular velocity as observed from the point of projection at the time of landing. = or
u sin R
=
g 2 u cos
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Component of velocity v perpendicu lar to line joining Objective : explain = = R Length of line joining S
Ex.
A spotlight S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/sec. The spot of light P moves along the wall at a distance of 3m. The velocity of the spot P when = 45° is _____ m/sec.
Ans:
V = 0.6m/s
Sol.
d =3 2
3m v
d= 3 2m Let velocity of spot at = 45° in V, then
45°
d
V
V d
P
S
Component of velocity perpendicu lar to line joining = Length of line joining
=
3m
45°
P
V 2 0.1 = 3 3 V = 0.3 2 × V = 0.6 m/s
2
Angular Acceleration: Rate of change angular velocity d dt –2 Unit rad s / Dimensions T–2
=
Direction is along angular velocity if it is increasing otherwise opposite. For uniform angular acceleration 1 2 2 t , = 20 + 2 2 0 = 10 rad s–1 and = –5 rad s–2 (uniform). Find angular displacement and number of turns at t = 6 sec.
= 0 + t , = 0t + Ex.
Sol.
1 ×5p ×36 = –30 2 Number of turns 25.
= 60 –
Objective : Explain why not 15 (students will give this as answer). Explain why distance is not equal to displacement. [Home Work : Read HCV Chapter-7)
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3.
KINEMATICS OF CIRCULAR MOTION:
NOTE : for XI class only (Derivation from HCV can also be used) Derivation of centripetal acceleration for a particle moving in a circle with constant speed. V2 = V1 = V
acceleration vector a , a =
V2 V1 t
V = V2 ( V1 ) V = 2 V sin 2
or V = Vfor small () V a =a= t
V2 =2R R
= V =
We can also prove that it is directed towards centre as angle with tangent is 90 in lim t 0. 2
lim 0 and angle 90° i.e. towards the centre aˆ Vˆ which is towards the centre. What we observe here is that when magnitude of velocity is constant but only direction is changing acceleration is directed perpendicular to velocity. In other words component of acceleration perpendicular to velocity causes change in direction not magnitude. Now lets take a case of particle moving in straight line, with changing speed. Its acceleration will be along the line of velocity. This acceleration will change only magnitude.
Imp.
Conclusion: Comp. of acceleration along velocity called tangential acceleration changes magnitude and comp. of acceleration perpendicular to velocity called radial acceleration changes direction of velocity.
Hence a a r a t . Derivation of tangential acceleration for circular motion. a=
dv dt
d ( R) dt
Rd dt
= R
=R
dv dt
=R
understand that v is speed or magnitude of velocity, v = v & rate of change of magnitude of velocity is called tangential acceleration. at = =
d V dt
= R.
or
dv dt
(a t R ) Page # 4
dV Also note that dt = a = a 2t a 2r d V dV i.e. dt = (R ) 2 (2 R 2 ) ( while = R.) dt Note : One is a magnitude of rate of change of velocity & other is rate of change of magnitude of velocity (at).
dv d|v| | a | | a | and t dt dt
Note : for XIII class only
d d v r = r dt dt dv d r d a r = ac a t dt dt dt Where , a c v and a t r 4.
CENTRIPETAL CONDITION: When component of acceleration perpendicular to velocity is
V2 R
and
always directed towards a fixed point then particle will undergo circular motion about that fixed point. ac =ar = Ex.
V2 R
If a particle is undergoing circular motion with speed V and radius R angle between acceleration & V is find magnitude of tangential acceleration in terms of V, R & . ac =
V2 R
= a sin
hence a = at=
V2 R sin
V 2 cos R sin
=
Objective :
V2 R
at = a cos
cot
at and ac are components of a and a & v may be at any angle between 0 to 180°
dV
Note:
dt
= a cos (Component of acceleration along velocity)
Valid for any type of motion (circle or not circle)
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Asking Question Is this diagram possible?
Ans.
This acceleration is not possible as perpendicular comp. of acceleration is away from curvature.
Asking Question Is this diagram possible?
Ans
This is possible but speed will be decreasing.
5.
RADIUS OF CURVATURE: For general curvilinear motion. When the particle crosses this point A, it is 2 satisfying condition of moving on this imaginary circle at this instant, if a = V / where Rc is radius of
Rc
curvature at this instant. Rc= Rc=
Ex. Sol.
V2 a
(speed)2 comp.of acceleration perpendicular to velocity
A projectile is launched horizontally with 20 ms–1 from some height. Find Rc at t = 2 sec. 20 m/s after t = 2 sec 20
gcos 45° 45° 20m/s
20 2
g
V2 RC = a
=
400 2 2 10
RC = 80 2 m
Objective : Explain Rc=
V2 a
=
(speed)2 comp.of acceleration perpendicular to velocity
Note: V , U Velocity / v, u speed V U at v u (a t ) t
but V U a t is wrong.
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CIRCULAR DYNAMICS INTRODUCTION: Component of Net force perpendicular to velocity should be inwards and its magnitude should be equal to
mV 2 R
. This is condition for circular motion.
STEPS FOR SOLVING PROBLEMS. Identify the plane of circular motion. Locate the centre and calculate the radius. Make F.B.D. Resolve forces only and always along these three directions: (a) In the plane along radial direction. (b) In the plane along tangential direction. (c) Perpendicular to the plane of circular motion. (a) Add the forces assuming radially inward direction as positive
Fr
=
mV 2 R
(b)
Ex.
Sol.
(c) If plane is not accelerating then F = 0. If plane is accelerating then actually it will not be circular motion from ground frame. But still F= ma , where a is acceleration of particle perpendicular to plane. Find tension in OA before and after AB is cut.
Before cutting a = 0 in all directions Revolving vertically & horizontally T1 cos = mg T1 sin = T1 T1 =
mg cos
Objective : This discussion is being done to explain what happens when we do not follow step (4) and resolve forces along other directions.
After cutting
T2 cos – mg = 0
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The above equation is wrong because acceleration of a particle in vertical direction is not zero. while T2 – mg cos = 0 is true as T2– mg cos = m 2 l (as = 0) What if we want to write equation in vertical direction. at = g sin / ac = 0 mg – T2 cos = m at sin 2 T2 cos = mg (1 – sin ) T2 = mg cos
Ex
Two different masses are connected to two light and inextensible strings as shown in the figure. Both masses rotate about a central fixed point with constant angular speed of 10 rad s–1 on a smooth horizontal plane. Find the ratio T1 of tensions T in the strings. 2 [Ans.
9 ] 8
Page # 8
Sol.
Drawing the FBDs for masses M1 and M2 T1 – T2 = M1R12
T2 = M2R22
T1 T2 M1 R1 1 1 · · = T2 M2 R 2 = 4 2
T1 1 9 = 1 + = T2 8 8
]
Objective : Centripetal force is the net force directed towards centre. Ex.
A car is moving in a circular path of radius 50 m, on a flat rough horizontal ground. The mass of the car is 1000 kg. At a certain moment, when the constant speed of the car is 5 m/s, find the force of friction acting on it? [Note : v gr for uniform speed]
[Ans.
500 N ]
Objective: Tell that only ext. force which can drive a car is friction. In case of car the direction of friction is not decided by velocity but according to need. Ex.
A car is moving in a circular path of radius 50 m, on a flat rough horizontal ground. The mass of the car is 1000 kg. At a certain moment, when the speed of the car is 5 m/s, the driver is increasing speed at the rate of 1 m/s2. Find the value of static friction on tyres (total) at this moment, in Newtons. 2
[Sol.
Fnet
2 v 2 dv 2 1 2 = m = m (1) r dt 2
=
m 5 = 500 5 N Ans. 2
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Objective : Friction is providing tangential as well as centripetal acceleration both
Ex.
A particle suspended from the ceiling by inextensible light string is moving along a horizontal circle of radius 0.05 m as shown. The string traces a cone of height 0.1 m. Find the speed.
0.1m
Conical pendulum 0.05m
[Sol.
mv 2 T sin = r
v=
;
g r tan = 0.5 m
& T cos = mg ]
Objective : E xpl ai n t he STEPS FOR SOLVING PROBLEMS also expalin why T cos = mg & not T = mg cos Ex.
In a rotor, a hollow vertical cyclindrical structure rotates about its axis and a person rests against the inner wall. At a particular speed of the rotor, the floor below the person is removed and the person hangs resting against the wall without any floor. If athe radius of the rotor is 2m and the coefficient of static friction between the wall and the person is 0.2, find the minimum speed at which the floor may be removed. Take g = 10 m/s2.
fs N
• mg
Ans. [Sol.
rg s =
2m 10m / s 2 v= = 10 m/s. 0. 2 Refer to HCV part-1 : Q.10 page-110 ]
Objective : N is not always equal to mg but any force directed towards centre is called centripetal force. [Home work : HCV (part-1) Ex. 1 – 17 ]
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Note : Give the following example for reading from HCV part-1 : Q.12 page-110 Ex. A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring. [Sol. Refer to HCV part-1 : Q.12 page-110 ] Sol. Consider a small part ACB of the ring that subtends an angle at the centre as shown in figure. Let the tension in the ring be T. T A O
/2 /2
C B
T
The forces on this small part ACB are (a) tension T by the part of the ring left to A, (b) tension T by the part of the ring right to B, (c) weight (m)g and (d) normal force N by the table. The tension at A acts along the tangent at A and the tension at B acts along the tangent at B. As the small part ACB moves in a circle of radius R at a constant speed v, its acceleration is towards the centre (along CO) and has a magnitude (m)v2 / R. Resolving the forces along the radius CO, + T cos 90 = (m) T cos 90 2 2
v2 R
v2 or, ... (i) R The length of the part ACB is R. As the total mass of the ring is m, the mass of the part ACB will be
2T sin = (m) 2
m m R = . 2R 2 Putting m in (i),
m =
v2 m 2T sin = R 2 2
or,
/ 2 mv 2 mv 2 T= 1 and T = 2 R sin ( / 2) 2R
]
Objective : Net force on the ring is zero but still it has tension. Q.
A block of mass m is sitting on a rotating frictionless wedge. The wedge rotates with constant angular velocity around the axis shown in figure.
Sol.
Calculate the value of such that the block stays at constant height h. (express your answer in terms of g, h,) h The velocity of P point on the wedge. Velocity of wedge at point P = r = × tan Drawing a free body diagram for the block. Page # 11
Resolving forces as indicated in step 2 .4 h h = tan =r r tan Ncos – mg = 0 (y–direction) 2 Nsin = m r 2 r 2 h tan = = g g tan
g tan 2 = h 2
=
g tan Ans.] h
Obejctive : Explain the concept of banking. Explain banking then take following as example [Refer HCV Ch.-7 (7.5) ] What is the speed required to negotiate the turn shown in figure. Frictionless and radius of curvature 'R' and banking ''. N cos O
v
N N sin C
R f mg
inside
N cos – mg = 0 N cos = mg N sin =
mv2 R
mg tan =
mv2 R
v = Rg tan Q.
A block of mass 25 kg rests on a horizontal floor ( = 0.2). It is attached by a 5m long horizontal rope to a peg fixed on floor. The block is pushed along the ground with an initial velocity of 10 m/s so that it moves in a circle around the peg. Find (a) Tangential acceleration of the block (b) Speed of the block at time t. (c) Time when tension in rope becomes zero. [Ans. (a) – 2 m/s2, (b) 10 – 2t, (c) 5 sec ] [Sol.(a) Tangential acceleration is the retardation produced by the friction a = –f/m = – mg/m at = – 0.2 × 10 = – 2 m/s2 Page # 12
(b)
dv = at = – 2 dt v
t
dv = 2 dt 10
(c)
0
v – 10 = – 2t v = 10 – 2t Tension in the rope will become zero when centripetal acceleration becomes zero i.e. when speed becomes zero v=0 10 – 2t = 0 t = 5 sec
Objective : Force directed towards centre is centripetal & force in the velocity direction is tangential force. Tangential acceleration is responsible for change in speed. If friction is kinetic it is opposite to velocity. If there is slipping on surface then net friction is kinetic. [Refer HCV Ch.-7 (7.6) ] CENTRIFUGAL FORCE: If reference frame is particle (undergoing circular motion) itself then it will experience pseudo force which will be radially outward and equal to m(w2R). Do not use it to explain any of the e.g. and let students solve all problems w/o it. For man outside N = mw2R
For man inside N – mw2R = 0. This is called centrifugal force.
Solved e.g. 13 H.C.V. Chap.7 take illustration in class without mentioning coriolis force {Home work : HCV 18 – 30 } Leave Effect of rotation of earth (to be discussed with gravitation.) Note : For any suggestion or correction please contact Amit Gupta or give it in computer room.
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