A small solid cylinder of radius r and mass M slides down a smooth hill of height h from rest and gets onto the plank of mass M lying on the smooth horizontal plane at the base of the hill as shown in the figure. Due to friction between the cylinder and plank, the cylinder slows down and starts rolling without friction over the plank. The coefficient of friction between the cylinder and plank is $μ$ .
Assume that height of the plank is negligible. Which of the following statement(s) is (are) true

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The velocity of center of mass of the cylinder after it starts pure rolling is $\frac{3}{4} \sqrt{2 g h}$

The velocity of center of mass of the plank – cylinder system when the cylinder is rolling over the plank is constant and equals to $\sqrt{\frac{g h}{2}}$

The minimum length of the plank required for pure rolling of the cylinder over the plank is $\frac{3 h}{8 μ}$

The fraction of initial mechanical energy that is lost due to friction is $\frac{3}{8}$

answer is A, B, C.

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Detailed Solution

$v_{C} - r ω = v_{P} M u = M v_{C} + M v_{P}$

Velocity of plank
$v_{P} = μ g t$ and $ω = \frac{2 μ g}{r} t$
Where $u = \sqrt{2 g h}$
$\Rightarrow v_{C} = \frac{3}{4} \sqrt{2 g h}; v_{P} = \frac{\sqrt{2 g h}}{4}; ω = \frac{\sqrt{2 g h}}{2 r} A s F_{e x t} = 0 ∴ v_{C m} = \frac{M \sqrt{2 g h}}{2 M} = \frac{2 g h}{2} = C o n s \tan t$

In the plank frame,

$L = \sqrt{2 g h t} - μ g t^{2} A s, t = \frac{v_{P}}{μ g} = \frac{\sqrt{2 g h}}{4 μ g} ∴ L = \frac{3}{16} \frac{u^{2}}{μ g} = \frac{3 h}{8 μ}$

Fraction of kinetic energy lost $= - \frac{W_{f r}}{t o t a l m e c h a n i c a l e n e r g y} = \frac{3}{8}$

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