Physics

Session

Rotational Mechanics - 6

Session Objectives

Session Objective

1. Angular momentum of a body in rolling

2. Friction in rolling

3. Motion of rigid bodies on frictionless and frictional inclined/horizontal plane

4. Problem Solving

Angular Momentum in Rolling

The angular momentum about point of contact (P) is given by

LP = mvcmr + Icw … (i)

Vcm

P

Friction in Forward Slipping

VCM

VCM > r

kf N CMV r

Friction in Backward Slipping

VCM

VCM < r

kf NCMr V

Friction in Rolling on an Inclined Plan

2

gsina

I1

MR

For rotational motion, f · R = I

For translational motion,

mg sin – f = maθ

For rolling,

a = R

R

f

mgsin

Class Test

Class Exercise - 1

A heavy disc is rolling down without slipping down an inclined plane, which is rough. Let V1 be its velocity at the bottom. Let V2 be the velocity of the same ball if the plane were smooth. Then

(a) V1 < V2 (b) V1 = V2

(c) V1 > V2 (d) Cannot be predicted

Solution

V2 > V1

Applying conservation of energy in thecase of pure rolling,

2 2

1 1 1 11 1

mgh mV I V r2 2

Applying conservation of energy in the case of sliding

22

1mgh mV

2

Hence, answer is (a).

Class Exercise - 2

A hollow cylinder rolls down a rough incline. Its velocity at the bottom is affected by

(a) radius (b) length

(c) density (d) change in height

Solution

a, c, d

Velocity will be affected by:

a. Radius — It changes the M.I.

b. Density — It changes the mass distribution and,hence, the M.I.

c. Change in height — It changes the total energy.

Class Exercise - 3A uniform circular disc of radius r placed on a rough horizontal plane has initial velocity V0 and an angular velocity w0 as shown. The disc comes to rest after some time. Then(a) friction acts in the backward direction

(b) the point of contact is at rest

0

0r

(c) V in magnitude2

(d) V0 = 2r0 in magnitude Voo

Solution

As the point of contact moves forward, the friction will act in the backward direction.

Hence, answer is (c).

Class Exercise - 4

A cylinder of mass m rolls down a rough inclined plane without slipping. Then which is/are correct?

(a) Acceleration is lower than that on a smooth plane

(b) Energy is lower than that on a smooth plane

(c) KE is related to the coefficient of friction ()

(d) The friction force may not be N, where N is the normal force

Solution

Acceleration will be lower than that ona smooth plane, as in pure rolling

2

gsina

I1

mR

Hence, answer is (a).

Class Exercise - 5

A solid sphere of mass M and radius R rolls down an inclined plane of inclination . The acceleration of the sphere down the plane is

2(a) g sin (b) gsin

35 7

(c) gsin (d) gsin7 5

Solution

22

2

gsin gsina

I 21 MR

5MR 1MR

5

gsin7

Hence, answer is (c).

Class Exercise - 6

Which of the following statements is not true?

(a) The instantaneous speed of the point of contact of disc rolling perfectly on a rough horizontal plane is zero

(b) The frictional force on the same body is zero

(c) The instantaneous acceleration of the point of contact of the same body is zero

(d) Work done against frictional force in the case of same body is not zero

Solution

b, d

The answers are definition of rolling.

Class Exercise - 7

Body A is placed on a rough horizontal surface with a horizontal speed of V0 and a spin of 0 as shown. Then

(a) it moves faster and its rotation slows down

(b) it moves slower and its rotation slows down

(c) there is no friction at P

(d) it stops

Voo

P

A

Solution

As tends to produce a forwardmotion at the point of contact P,friction is in backward direction.So both V0 and reduce.

0

0

Hence, answer is (b).

Friction Translation dueto rolling inforward direction

Vo

o

P

Class Exercise - 8

A ball is released on a rough horizontal plane with a speed v0 so that it slides on the surface without rolling. To what value should the speed decrease so that it rolls without sliding?

Vo

P

Solution

The moment the ball touches thesurface with speed v0 to right,frictional force f acts to the left.

Torque due to f producesclockwise rotation (so increases from initial zerovalue) while f(= mmg)produces a reduction of v0.

Let the pure rolling motionstart at linear velocity v.

( v = r )

Vo

P

r

f

Solution contd..

Translation:

Acceleration

mg

g (m : mass of ball)m

0v v gt

Rotation:

0 t

2

2mgr 5 mgr 2

I mrI I 52mr

5 g 5 gt

2r 2r

Solution contd..

0 0v vv gt 5 gt gt

Alsor r r 2r r r

0 0v 2v7 gt

t2r r 7 g

0

0 0g 2v 5

So, v v v v7 g 7

Class Exercise - 9

A solid sphere (radius R, mass M) is given a clockwise angular velocity 0 and lowered to a flat horizontal surface with coefficient of friction with the sphere. The sphere first slips on the surface. After time t the motion changes to pure rolling without slipping. What is the value of t and the value of velocity of centre of mass at t?

Solution

When rolling starts, let v and be thelinear and angular velocities respectively.F = Mg is the friction force (inthe forward direction), whichproduces an acceleration a = g.

v

o

f

Solution

v 0 gt ...(i)

The torque is fR = Mg R and

2MgR 2

I MRI 5

5 g

2R

0

5 gt ...(ii)

2R

0

gt 5 g(i) and (ii) give t

R 2R

0

02 R7 gt

t2 R 7 g

02

v R7

Class Exercise - 10

A disc of radius b and mass m rolls down an inclined plane of vertical height h. The translational speed when it reaches the bottom of the plane will be

4gh(a) (b) 2gh

3

7gh 5gh(c) (d)

3 7

Solution

Applying the conservation of energy,

2 21 1mgh mv I

2 2

2 22 2

21 1 mb v 3

mv mv2 2 2 4b

4gh

v3

Hence, answer is (a).

Thank you