05 Mechanics

ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ MECHANICS KINEMATICS :

Motion on a Straight Line (one dimensional motion)

Uniform velocity i) s = vt ii) a = 0

Motion with constant acceleration u+v i) s =  t  2  1 ii) s = ut + at 2 2 2 2 iii) v = u + 2as iv) v = u + at

Motion with variable acceleration dv i) If a = f (t), a = dt dv ii) If a = f (s), a = v ds dv iii) If a = f (v), a = dt ds iv) v = dt v) s = ∫ vdt vi) v = ∫ adt

1.

Projectile :

gx 2 . 2u 2 cos 2 θ b) The equation of trajectory of projectile is a parabola with y 2 2 2u cos θ i) Latus rectum = g O  u 2 sin θ cos θ u 2 sin 2 θ  ii) The coordinates of the focus =  ,−  g 2g   2 u iii) The equation of directrix, y = . 2g

a) The equation of trajectory of projectile is y = x tan θ −

u 2 sin 2θ 2u sin θ , Range R = g g ˆ d) Velocity at any time t, v = u cos θi + (u sin θ − gt)ˆj . 1   e) Position at any time t, r = u cos θ t î +  u sin θt − gt 2  ˆj  2 

u θ

x

c) Time of flight =

f) Range of projectile on inclined plane R = g) Time of flight T =

u2 [sin(2θ − β) − sin β] . g cos 2 β

2u sin(θ − β) g cos β

h) Maximum Range =

u2 . g(1 + sin β)

ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ

u θ

β

ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ 2.

Motion along a curve : Radial and Transverse components : y

ˆj

v vt

Radial

eˆ r

r

θ

θ

î Position vector r = reˆ r

x

eˆ r = cos θî + sin θˆj ; eˆ θ = − sin θî + cos θˆj

Where eˆ r = unit vector along radial direction, eˆ θ = unit vector along transverse direction. dr deˆ d v = (reˆ r ) = eˆ r + r r dt dt dt dr v = eˆ r + rωeˆ θ = r eˆ r + rωeˆ θ dt dv Acceleration a = dt dω a = (r eˆ r + rω eˆ θ ) + rω eˆ θ + reˆ θ − rω2eˆ r dt

= (r − rω2 )eˆ r + (2rω + rd)eˆ θ

{

}

1d 2 a = (r − rω2 )eˆ r + (r ω) eˆ θ r dt where r =

dr d2r , r = 2 . dt dt

DYNAMICS : m(v − u) change in momentum 1. Force = . , i.e., F = t time 2.

A jet of water of density d from a tube of area of cross section a comes out with a velocity v. Average force exerted by tube on water is dAv2.

3.

The impulse of a force is defined as the product of the average force and the time interval for which it acts. Impulse J = FAV ∆ t = mv − mu t2 Impulse momentum theorem = ∫ F dt = mv − mu t1

4.

A rope of length ‘L’ is pulled by a constant force ‘F’. The tension in the rope at a distance ‘x’ from the x end where it is applied is F(1 − ). L

5.

The amount of work done is given by the dot product of force and displacement. F.s = Fs cos θ If the force is varying non–uniformly, then the work done = ∫ F.ds = ∫ F.ds.cos θ .

6.


F

s

ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ 7. 8.

1 The work done on a spring in stretching or compressing it through a distance x is given W = kx2 2 where k is the force constant or spring constant. Work Energy Theorem : Work done on a body can produce a change in its kinetic energy. According to work energy theorem, change in kinetic energy of a system is equal to work done by all the forces (internal forces, external forces, conservative forces and non-conservative forces) on the system. W = ∆K = K f − K i

9.

Conservative Force : The force which does work in complete independence of the path followed by the body is called a conservative force. The gravitational force, spring force and electrostatic force are the examples of the conservative forces. 10. Non-conservative Force : The work done by a non-conservative force not only depends upon the initial and final positions but also on the path followed. The common examples of such forces are frictional force and drag force of fluids. 11. Potential Energy : The potential energy is defined only for conservative forces. B

U B − U A = − ∫ Fc ⋅ ds A

12. Conservative force : U(x) A

C

B

At equilibrium,

x

dU =0 dx

The point B is the position of stable equilibrium, because

d2U >0. dx 2

d2U < 0. dx 2 13. In the absence of any external and nonconservative forces (or even, if they are present mechanical energy (sum of kinetic and potential energies) of the system remains conserved. 14. Rate of doing work is called power. work = Force × velocity. Power = time dθ = τω 15. If a body is rotated in circular path, the power exerted is given by P = τ dt The point C is the position of unstable equilibrium, because

16. When water is coming out from a hose pipe of area of cross section 'A' with a velocity 'v' and hits a wall normally and i) stops dead, then force exerted by the water on the wall is Av2 ρ. And the power exerted by water is P = A v3 ρ (ρ = density of water) ii) If water rebounds with same velocity (v) after striking the wall, P = 2Av3 ρ. COLLISION : 1. Collision is an interaction between two or more bodies in which sudden changes of momenta take place. e.g. : Striking a ball with a bat.



3.

Coefficient of restitution (e) : The coefficient of restitution between two bodies in a collision is defined as the ratio of the relative velocity of separation after collision to the relative velocity of their approach before their collision. relative velocity of separation v −v ; (ii) e = 2 1 (i) e = relative velocity of approach u1 − u 2

m1 u1

A

a) V1 = u1 V2 = u1

m1

m2 u 2 B

A

v1

m2 v2 B

( m1 − em 2 ) + u (1 + e ) m2 m1 + m 2

2

m1 + m 2

m1 (1 + e ) ( m 2 − em1 ) + u2 m1 + m 2 m1 + m 2

V1 & V2 are velocities of m1 and m2 after the collision b) Loss of Kinetic energy from the system is ∆K = Elost =

1  m1m 2  2 2   [1 − e ] [u1 − u 2 ] 2  m1 + m 2 

c) e = 1 for perfectly elastic collision, e = 0 for perfectly inelastic collision. CENTRE OF MASS : 1. Every particle is attracted towards the centre of the earth by the force of gravity and the centre of gravity of a body is the point where the resultant force of attraction of the weight of the body acts.

2.

Centre of mass : If a system of parallel forces proportional to the masses of the various particles of a body are assumed to act on it, their resultant passes through a fixed point, irrespective of the direction of the parallel forces and that point is called centre of mass.

3.

In vector notation, each particle of the system can be described by a position vector r and the centre of mass can be located by the position vector rcm . mr ˆ rcm = ∑ i i where ri = îx i + ˆjyi + kz i ∑ mi ˆ rcm = îx cm + ˆjycm + kz cm

4.

Centre of Mass of Some Common Systems : i) A system of two point masses

The centre of mass lies closer to the heavier mass. L m2

m1

m2L m1L m1 + m 2 m1 + m 2

ii) A circular cone yc =

CM

h 4

h

yc


ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ iii)A semi-circular ring : yc =

2R ; xc = 0 π

Y

y

CM

R

X

O

iv) A semi-circular disc : yc =

4R ; xc = 0 3π

Y

y

CM

R

X

O

v) A hemispherical shell : yc =

R ; xc = 0 2

Y

y

CM

R

X O

vi) A solid hemisphere : yc =

3R ; xc = 0 8

Y

y

CM

R

X O


ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ FRICTION :

1.

When a body is in motion over another surface or when an object moves through a viscous medium like air or water or when a body rolls over another, there is a resistance to the motion because of the interaction of the object with its surroundings. Such a resistance force is called force of friction.

2.

Motion on a rough horizontal plane :

(a) Frictional force = µ × normal reaction. body moving with uniform velocity F = µ k mg. (b) Pulled with a force F inclined at an angle θ with the horizontal and the body moving with uniform velocity. F=

µ k mg cos θ + µ k sin θ

(c) Pushed with a force F inclined at an angle θ with the horizontal and the body moving with uniform velocity: F= 3.

µ k mg cos θ − µ k sin θ

Block on a rough inclined plane

a) Angle of repose α : It is the angle of inclination of the inclined plane with the horizontal for which block just begins to slide down. b) If α is the angle of repose µs = tanα c) The angle of repose is the angle of static friction d) The angle of inclination is (θ) less than (α), the block does not slide down, it is at rest. The force of friction f < fs and is equal to

N mg sin θ θ mg cos θ

θ mg

f = mg sinθ [mg sinθ < fs] e) If the angle of inclination is [θ] equal to [α]. Then the block is in limiting equilibrium. The force of friction is f = fs = µs mg cos α f) If the block slides down the inclined plane with uniform velocity µk = tan θ where θ is the angle of inclination of the inclined plane.


ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ ROTATORY MOTION : 1. Equations of motion of rotating body with constant angular acceleration : ω = ω0 + αt

1 θ = ω0 t + αt 2 2 2 2 ω = ω0 + 2αθ α (2n − 1) 2 Moment of inertia or rotational inertia : M.I. of a particle of mass m at a distance r from axis of rotation is given by I = mr 2 . M.I. of a body can be defined I = ∑ mi ri2 θn = ω0 +

2.

For a system of discrete particles I = ∫ r 2 dm . 3.

Radius of gyration : Radius of gyration of system of particles about an axis is given by K =

where I = ∑ mi ri2 and M = ∑ mi . Radius of gyration of a body of mass M and moment of inertia about an axis is given by K = 4.

Rolling motion : i) For a wheel of radius R rolling with constant velocity vcm = ωR .

ii) For rolling with constant acceleration a cm = αR . 5.

Velocity of different points of a rolling wheel : 2vcm 2 vcm

ω 2 vcm

Angular momentum of a rolling wheel : L = LTranslation + L Rotation or L = m(R × vcm ) + Icm ω . Kinetic energy of a rolling wheel : 1 1 1 k2  2 2 2 K.E. = K Translation + K Rotation = mvcm + Icm ω = mv 1 + 2  . 2 2 2  r  Rolling on rough inclined plane : Acceleration of C.M.  g sin θ  a cm =  I   1+   mR 2  2gh Velocity of C.M. after falling height h : vcm = I 1+ mR 2 mg sin θ Minimum friction required to cause pure rolling : f min = r2 1+ 2 k O

6. 7.

8.


I . M

I , M


Rolling of cylinder on inclined plane : 2g mg tan θ . a cm = sin θ , f min = sin θ and µ min = 3 3 3

GRAVITATION : 1. If m1 and m2 are the masses of two bodies and d is the distance between them, the gravitational force of attraction F which each exerts on the other is given by

F = G⋅

m1m 2 d

2

where G is called universal gravitational constant and is equal to 6.67×10−11 Nm2kg−2. GM

=

4 πRρG 3

where M is the mass of the planet, R is its

2.

The relation between g and G is given by g =

3. 4.

radius and ρ is the mean density of the planet. Variation of g due to rotation of the earth is given by g1 = g − R ω 2 cos 2 λ where λ = latitude angle. If g is the acceleration due to gravity on the surface of earth and gh at a height h above the earth, then gh =g(1−

R

2

2h gR 2 exactly. ) approximately or gh = R (R + h) 2 d R

).

5.

If d is the depth below the surface, then gd = g(1−

6.

Gravitational field strength is the force experienced by unit mass in the gravitational field. It is a vector.

Its magnitude is equal to the value of g. I = 7.

R2

Gravitational potential is the work done in moving a unit mass from infinity to the point under

consideration. V = − 8.

GM

GM R

or Ι = −

dV dR

The binding energy of a mass ‘m’ at rest on the surface of earth of mass M and radius R is given by U = −

GMm = − mgR . R

Generally it is stated as positive energy i.e., as mgR. 9.

Gravitational potential due to a mass M at a distance r is −

GM . r

GM = gR R 11. A satellite of mass m orbiting close to the earth has kinetic energy and potential energy.

10. For an orbit close to earth surface orbital velocity is V0 = GMm mgR = 2R 2 GMm 13. Potential energy of the satellite = − R GMm 14. Total energy = K.E + P.E = − (negative 2R

12. Kinetic energy of the satellite =

sign signifies that the body is bound to the earth)

15. The escape velocity of a body on earth or on any planet is Ve = 2gR or Ve =


2GM R

ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ SIMPLE HARMONIC MOTION :

1.

2. 3.

SHM : If the restoring force/torque acting on the bob in oscillatory motion is directly proportional to the displacement of body/particle and is always directed towards equilibrium position, then the motion is called Simple Harmonic Motion (SHM). F = –kx General equation of SHM is x = Asin(ωt + φ) ; (ωt + φ) in phase of the motion and φ is initial phase of the motion. 2π = 2πf Angular Frequency (ω) : ω = T

Time period (T) : T =

2π m = 2π ω k

k m

4.

Speed : v = ω A 2 − x 2

5.

Acceleration : a = −ω2 x 1 1 1 Kinetic Energy (KE) : mv 2 = mω2 (A 2 − x 2 ) = k(A 2 − x 2 ) 2 2 2 1 Potential Energy (PE) : kx 2 2 1 1 Total Mechanical Energy (TME) : KE + PE = k(A 2 − x 2 ) + kx 2 2 2 1 = kA 2 (which is constant). 2 Spring Mass System :

6. 7. 8.

9.

i)

k

k

⇒

m

m

m k

⇒ T = 2π

ii)

k m1

T = 2π

m2

µ m1m 2 known as reduced mass. where µ = m1 + m 2 k

l 1 = 2π (in accelerating reference frame) ; geff is net acceleration due g g eff to pseudo force and gravitational force. 11. Compound Pendulum / Physical Pendulum

10. Simple Pendulum : T = 2π

Time Period (T) : T = 2π

I , where I = ICM + ml2 ; l is distance between point of suspension and mgl

centre of mass.


ǁǁǁ͘ƐĂŬƐŚŝĞĚƵĐĂƚŝŽŶ͘ĐŽŵ Previous Questions : 1. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is (IIT-2010) P

2.

2GM (4 2 − 5) 7R GM 4R

4R

2GM (4 2 − 5) 7R 2GM ( 2 − 1) 5R

*a)

b) −

c)

d)

3R

A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is µ and tan θ > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sin θ – µ cosθ) to P2 = mg (sin θ + µ cos θ), the frictional force f versus P graph will look like (IIT-2010) *a)f b) f c) f d) f P2 P1

P

P1

P2

P1

P

P2

P1

P

P2

3.

4R

P θ

P

A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is(are) correct for the system of these two masses? *a) Total momentum of the system is 3 kg ms–1 b) Momentum of 5 kg mass after collision is 4 kg ms–1 *c) Kinetic energy of the centre of mass is 0.75 J d) Total kinetic energy of the system is 4 J 4. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? (IIT-2010) *a) Error ∆T in measuring T, the time period is 0.05 seconds b) Error ∆T in measuring T, the time period is 1 second *c) Percentage error in the determination of g is 5% d) Percentage error in the determination of g is 2.5% Passage : When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to m / k , as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = αx4 (α > 0) for |x| near the origin and becomes a constant equal to V0 for |x| ≥ X0 (see figure). 5. If the total energy of the particle is E, it will perform periodic motion only if d) E > V0 a) E < 0 b) E > 0 *c) V0 > E > 0 6. For periodic motion of small amplitude A, the time period T of this particle is proportional to a) A 7.

m α

*b)

1 m A α

c)

1 α A m

d)

1 α A m

The acceleration of this particle for |x| > X0 is a) proportional to V0 b) proportional to

V0 mX0

c) proportional to

V0 mX0


*d) zero


A binary star consists of two stars A (mass 2.2 MS) and B (mass 11 MS), where MS is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the (6) centre of mass is …. (IIT-2010) Sol.

9.

Gravitational acceleration on the surface of a planet is

6 11

g, where g is the gravitational acceleration

on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1 will be… (IIT-2010) Sol. (3) 10. A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The F(t) kinetic energy of the block after 4.5 seconds is (IIT-2010) 4N a) 4.50 J 4.5s b) 7.50 J 3s O *c) 5.06 J d) 14.06 J 11. A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is (IIT-2010) a) 0.02 mm b) 0.05 mm c) 0.1 mm *d) 0.2 mm 12. A uniform rectangular lamina ABCD of length and breadth ‘b’ is kept on a smooth flat floor. At one corner B it is pivoted to the floor so that it can rotate freely B about it. The other corner of the length is attached to an unstretched spring of A force b constant K, as shown in figure. The time period of small oscillations of the lamina about the pivot B. a) 2π

M( 2 + b 2 ) 2K 2

b) 2π

M( 2 + b 2 ) K 2

*c) 2π

M( 2 + b 2 ) 3K 2


d) 2π

M( 2 + b 2 ) 6K 2

05 Mechanics

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