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A uniform disc of mass M and radius R is rotating about its center of mass (the center of mass is at rest) with an angular speed $ω$ . The angular momentum of disc about a point A (as shown) will be

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$M R^{2} ω + M h R ω$

$\frac{1}{2} M R^{2} ω$

$\frac{1}{2} M R^{2} ω + M h R ω$

$M R^{2} ω$

answer is B.

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Detailed Solution

The angular momentum of disc about point A is ${\vec{L}}_{A} = I_{c m} \vec{ω} + m \vec{r} \times {\vec{v}}_{c m}$

${\vec{v}}_{c m} =$ Velocity of centre of mass of disc = 0.

$∴ L = I_{c m} ω = \frac{1}{2} M R^{2} ω$

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