JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Rotational Mechanics

A uniform thin cylindrical disk of mass

M

and radius

R

is attached to two identical massless springs of spring constant

k

which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance

d

from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is

L

. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance

L

from the wall. The disk rolls without slipping with velocity

V_{0} = V_{0} \hat{i}

. The coefficient of friction is

μ

.

The net external force acting on the disk when its centre of mass is at displacement

x

with respect to its equilibrium position is

Answer: (D)

Solution

When the disk is displaced by

x

, both springs exert a restoring force. The total spring force is

F_{s p r in g} = - 2 k x

.

For rolling without slipping, we have the constraint

a = R α

, where

a

is linear acceleration and

α

is angular acceleration.

Applying Newton's second law for translation:

2 k x - f = M a

Applying Newton's second law for rotation about the center:

f R = I α = \frac{1}{2} M R^{2} \cdot \frac{a}{R} = \frac{1}{2} M R a

From the rotational equation:

f = \frac{M a}{2}

Substituting into the translational equation:

2 k x - \frac{M a}{2} = M a

2 k x = \frac{3 M a}{2}

M a = \frac{4 k x}{3}

Therefore, the net external force is

F_{n e t} = M a = - \frac{4 k x}{3}

(negative for restoring force).

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option