QJEEAdv08Phe496: JEE Advanced 2008 Physics - Class 11 JEEAdv08Phe496 Solution: The maximum value of V for which the disk will roll without…

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

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

$\Rightarrow - 2kx + f = - 2f$

$\Rightarrow f = {{2k} \over 3} \times x$

Frictional force depends on x. As x increase friction increase. Frictional force (f) is maximum at $x = A$ .

Where A = amplitude of S.H.M

Maximum frictional force,

${f_{\max }} = {{2k} \over 3} \times A$

The force should be almost equal to the limiting friction ( $\mu mg$ ) for rolling without slipping.

$\therefore$ $\mu mg = {{2k} \over 3} \times A$ ..... (iv)

For S.H.M

Velocity amplitude $= A\omega$

$\therefore$ ${V_0} = A\omega$

$\therefore$ ${V_0} = {{3m\,mg} \over {2k}}w$ (from (iv))

$\therefore$ ${V_0} = {{3\mu mg} \over {2k}} \times \sqrt {{{4k} \over {3M}}}$ (As $\omega = \sqrt {{{4k} \over {3M}}}$ )

$\therefore$ ${V_0} = mg\sqrt {{{3M} \over k}}$