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 maximum value of

V_{0}

for which the disk will roll without slipping is

Answer: (C)

Solution

For rolling without slipping, the friction force must not exceed

f_{ma x} = μ M g

.

At maximum displacement

x_{ma x}

, the friction is maximum. From the force equation:

2 k x_{ma x} - f_{ma x} = M a_{ma x}

From the constraint that friction is maximum:

f_{ma x} = μ M g

From the rotational equation:

f_{ma x} = \frac{M a _{ma x}}{2}

Therefore:

μ M g = \frac{M a _{ma x}}{2}

, giving

a_{ma x} = 2 μg

Substituting back:

2 k x_{ma x} - μ M g = M (2 μg)

2 k x_{ma x} = 3 μ M g

x_{ma x} = \frac{3 μ M g}{2 k}

Using energy conservation:

\frac{1}{2} I_{P} ω_{0}^{2} = \frac{1}{2} (2 k) x_{ma x}^{2}

For a disk rolling:

I_{P} = \frac{3 M R ^{2}}{2}

and

ω_{0} = \frac{V _{0}}{R}

\frac{1}{2} \cdot \frac{3 M R ^{2}}{2} \cdot \frac{V _{0}^{2}}{R ^{2}} = k (\frac{3 μ M g}{2 k})^{2}

\frac{3 M V _{0}^{2}}{4} = \frac{9 μ ^{2} M ^{2} g ^{2}}{4 k}

V_{0} = μg \frac{3 M}{k}

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