First slide

Rolling motion

Question

A solid sphere rests on a horizontal surface. A horizontal impulse is applied at height h from centre. The sphere starts rolling just after the application of impulse. The
ratio h/r will be

Difficult

A

$\frac{1}{2}$
B

$\frac{2}{5}$
C

$\frac{1}{5}$
D

$\frac{2}{3}$

Solution

Rolling is rotation about point of contact. Applying impulse momentum equation about P.
$\begin{array}{l} J (R + h) = I_{P} ω . . . . . (i) \\ and J = m v . . . . . . (i i) \end{array}$
$As sphere rolls v = ω R, and I = \frac{2}{5} m R^{2} + m R^{2} = \frac{7}{5} m R^{2}$
$After solving, we get \frac{h}{R} = \frac{2}{5}$

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