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A disc of mass m and radius R is rolling without slipping on a horizontal plane with angular velocity $ω .$ P is a point in space in the plane of the disc at a height 2R above the horizontal plane. Then angular momentum of the disc about point P is

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$\frac{1}{4} ω m R^{2}$

$ω m R^{2}$

$2 ω m R^{2}$

$\frac{1}{2} ω m R^{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} {\vec{L}}_{P} = {({\vec{L}}_{c o m})}_{P} + I_{C} \vec{ω} \\ L_{P} = m V_{c} R - I_{C} . ω = m (ω R) R - \frac{1}{2} m R^{2} ω = \frac{1}{2} m ω R^{2} \end{array}$

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