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JEE Advance - Physics (2008 - Paper 2 Offline - No. 18)

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 $${\overrightarrow V _0} = {V_0}\widehat i$$. the coefficient of friction is $$\mu$$.

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 $${\overrightarrow V _0} = {V_0}\widehat i$$. the coefficient of friction is $$\mu$$.

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 $${\overrightarrow V _0} = {V_0}\widehat i$$. the coefficient of friction is $$\mu$$.

The centre of mass of the disk undergoes simple harmonic motion with angular frequency $$\omega$$ equal to:
$$\sqrt {{k \over M}} $$
$$\sqrt {{{2k} \over M}} $$
$$\sqrt {{{2k} \over {3M}}} $$
$$\sqrt {{{4k} \over {3M}}} $$

Explanation

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

IIT-JEE 2008 Paper 2 Offline Physics - Rotational Motion Question 14 English Explanation

$$\therefore -2kx+f=-M_{ac}$$ ..... (i)

Where $$a_c$$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$$\tau = f \times R = I \times {a_c}$$

$$\therefore$$ $$f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$$

[$$\therefore$$ $$I = {1 \over 2}M{R^2},{a_c} = R{a_c}$$ for rolling without slipping]

$$\therefore$$ $$f = {1 \over 2}M{a_c}$$ ...... (ii)

From eq. (i) and (ii),

$$ - 2kx + {1 \over 2}M{a_c} = - M{a_c}$$

$$ \Rightarrow {3 \over 2}M{a_c} = 2kx$$

$$ \Rightarrow M{a_c} = {{4kx} \over 3}$$

$$\Rightarrow$$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $$ = {{ - 4kx} \over 3}$$ directed towards the equilibrium.

$$ \therefore $$ $$|{F_{net}}| = {{4k} \over 3}x$$

For S.H.M $$|{F_{net}}| = M{\omega ^2}x$$

$$\therefore$$ $$M{\omega ^2} = {{4k} \over 3}$$

$$ \Rightarrow \omega = \sqrt {{{4k} \over {3M}}} $$ .... (iii)

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