Physics

Session

Rotational Mechanics - 5

Session Objectives

Session Objective

1. Rolling of a body – Description

2. Rolling of a body - Mathematical expression

3. Instantaneous center of rotation

Rolling of a body

Rolling without slipping : Motion is combined rotation and translation.

Point of contact has no relative motion with surface

There is no relative motion between body and surface

VP

Q surface

Rolling of a body

Friction (f) is responsible for rolling without slipping

No work is done by f when body rolls without slipping

f = 0slipping

21KE MV

2

spinningno translation

21KE I

2

P

f 0rolling

2 2cm

1 1KE MV I

2 2

Relationship between linear & angular velocity

The point in contact with surface is momentarily at rest. Linear velocity is velocity of center of mass (o).

Or P Vcm

Linear distance moved by centre of mass = S

It is also the distance PQ along the edge of circular body

VCMr

r

S Q

P

Relationship between linear & angular velocity

S r (r : radius of circular cross section)

CM

ds dv r r

dt dt

CMV r

Dynamics of rolling

Rolling without slipping = Linear motion of center of mass+ circular motion about center of mass.

Point P is at rest with respect to the surface at this instant.It is the instantaneous center of rotation .

VCM

VCM

VCM

P

2VCM

VCM

V = 0

+ O

rr

r r

Kinetic Energy

When considered with respect to P

2 2p p CM

1KE I But I I Mr

2

2 2CM CM

1 1KE I MV

2 2

CM( v r ) 777777777777777777777777777777777777777777

2 CMCM 2

I1v M

2 r

O VCM

P

Q

pr

r

r

Kinetic Energy with respect to instantaneous axis of rotation

Velocity vQ of the point Q, with respect to the instantaneous centre rotation

p CM pv v r r 777777777777777777777777777777777777777777777777777777777777777777777777777777777777

ppand a r 7777777777777 7

So torque (external) = (I measured from P)

I

21KE I and L I

2

7777777777777 7

are valid about instantaneous point of rotation.

O VCM

P

pr

r

r

pV77777777777777

Class Test

Class Exercise - 1

A disc, a ring and a sphere of same radius and same mass are placed on a rough horizontal table so that they can roll over the table. An equal impulsive force is imparted to each through their centres of mass and they start rolling and slipping. When the slipping stops the linear velocity is greatest for

(a) ring (b) disc

(c) sphere (d) All are equal

Solution

2 2 22

1 1 1 IKE I mv v m

2 2 2 r

Ring: I = mr2 2 2r

1KE v (2m) mv

2

21Disc : I mr

2

2 2d

1 1 3KE v m m mv

2 2 4

22Sphere : I mr

5

2 2s

1 2 7KE v m m mv

2 5 10

For same mass and KE, sphere has maximum linearvelocity.

Hence, answer is (c).

Class Exercise - 2

A solid sphere rolls on a horizontal plane without slipping, percentage of KE which is rotational is approximately

(a) 28 (b) 3

(c) 72 (d) 100

Solution

For a sphere,

2 2

2 2 2

2

1 1KE I mv

2 2

1 2mr mv

2 5

1mv (0.4 1)

20.4

Rotational KE percentage 100%1.428%

Hence, answer is (a).

Class Exercise - 3

A hoop of radius r rolls over a horizontal plane with constant velocity v without slipping. The velocity of any point t seconds after it passes the top position is

v

vt vt(a) v cos (b) 2v cos

r rvt vt

(c) 2v sin (d) 2v cosr 2r

Solution

is constant.

P

r

B

C

r

r´

2

–

2–

2 V

v´

v

A

Speed at B = v’

v ' r ', v r

1 cosr

cos2

Solution contd..

2

r r cos r(1 cosr '

cos(CPB) cos2

r 1 2cos 12

2r cos2cos

2

v ' 2 r cos2t

2v cos t2vt

2v cos2r

Hence, answer is (d).

Class Exercise - 4

A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and moment of inertia about its axis is I . A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

2

2

2gh 2mgh(a) r (b) r

mr mr

2 mgh(c) (d) 2gh

2mr

Solution

v = r

Loss in PE = Gain in KE

m

v

h

m

Solution contd..

2 2wheel

22

2

22 2

2 2

2

1 1mgh I mv

2 2

1 v 1I mv

2 2r

1 I 1 I mrm v v

2 2r r

2mghv r

I mr

Hence, answer is (b).

Class Exercise - 5

A mass M is supported by a massless string wound round a uniform cylinder of mass M and radius R . On releasing the system from rest, the acceleration of mass M is

R

M

(a) g (b)

(c) (d) depends on R

g2

2g

3

Solution

v = r

M

R M Loss in PE = Gain in KEwhen M has dropped a height h.

Solution contd..

2 2

22 2

2

2

2

1 1Mgh Mv I

2 2

1 1 1 vMv Mr

2 2 2 r3

gh v4

3vh

4g

2v 0 2ah 2v 2

a g2h 3

Hence, answer is (c).

Class Exercise - 6

A ring and a cylinder of the same mass m and the same radius R are released with the same velocity at the same time on a flat horizontal surface such that they start rolling immediately on release towards a wall equidistant from both. The rolling friction is finite. Which of the two will reach the wall first?

Solution

For rolling,

2 21 1KE mv I [v = r ]

2 2

2

21 I

mv 12 mr

As m and v are same, KE is same for both.For the ring: I = mr2

2r

KEmv KE v

m

Solution contd..

21For the cylinder : I mr

2

2c

3 4 KEKE mv v

4 3 m

c rv v

So the cylinder will reach first.

Class Exercise - 7

A body of mass m and radius r is rolled up on irregular surface with finite but negligible friction so that it attains a

maximum vertical height of .

What is the moment of inertia of the body and what may be its shape?

25 v

6 g

v

h

Solution

2

2 2 22

1 1 1 1 vKE mv I mv I

2 2 2 2 r

v for rolling

r

2

21 I

v m2 r

As friction is negligible,Loss in KE = Gain in PE

25 mgv

mgh6 g

22

2

5 mgv 1 Iv m 1

6 g 2 mr

2

5 II

3 mr 22

I mr3

The body is probably a hollowsphere.

Solution contd..

Hint/Formula:

2

21 I

mv 1 mgh2 mr

2 2 2

21 v I 1 v 2 5 v

h 1 12 g 2 g 3 6 gmr

Class exercise - 8A cylinder AB of radius R, length L and mass M has two cords A and B wound around it. The ends of the cords are attached to a fixed, horizontal support and the cords are vertical. When the cylinder is released, the cords unwind and the cylinder moves down while rotating. What is the angular acceleration of the cylinder and the tension in the cords as the cylinder falls?

R

A B

L

Solution

The equations of motion:

Mg – 2T = Ma (Translational)

2TR = I (Rotational)

2MR

I2

and (Motion of cylinder is rotational) a

R

21 a 1Then 2TR MR MRa

2 R 2

1

2T Ma2

1

Mg Ma Ma2

2 2g

a g3 3R

1

T Mg6

Class Exercise - 9

A disc of radius R is spin about its axis so that its angular velocity is w. Then it is slowly placed on a rough horizontal surface of coefficient of friction m. How long will the disc takes to stop? (Plane of the disc is parallel to surface all through the rotation.)

Solution

Take a concentric ring of radius r andwidth dr (as one of a series ofconcentric rings making up the disc).

rdr

Its mass: 2 2M M

dM 2 rdr 2rdrR R

Solution contd..

Torque on the ring due to friction:

d = dMrg

So total torque on the disc opposing the motion:

R R

22

0 0

2 Mg 2d r dr MgR

3R

d 2I MgR

dt 3

21

I MR2

t 0

0

3 Rdt d

4 g

3 Rt

4 g

Thank you