The disc starts rotating anticlockwise with the same angular velocity ω=vR, then what will be the angular momentum of the disc about bottommost point in this new situation?

L0=mvR-Icω=mvR-12mR2(vR) =12mvR ( clockwise )

The disc starts rotating anticlockwise with the same angular velocity ω=vR, then what will be the angular momentum of the disc about bottommost point in this new situation?

L0=mvR-Icω=mvR-12mR2(vR) =12mvR ( clockwise )

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The disc starts rotating anticlockwise with the same angular velocity $ω = \frac{v}{R}$ , then what will be the angular momentum of the disc about bottommost point in this new situation?

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$\frac{2}{m} vR$

$\frac{1}{2} mvR$

$\frac{1}{5} mvR$

$\frac{1}{3} mvR$

answer is D.

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Detailed Solution

$L_{0} = mvR - I_{c} ω$

A solid sphere of mass in and radius R is rolling without slipping as shown in the figure. Suppose the disc starts rotating anticlockwise with the same and velocity omega =frac{v}{R}, then

$= mvR - (\frac{1}{2} {mR}^{2}) (vR) = \frac{1}{2} mvR (clockwise)$

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The disc starts rotating anticlockwise with the same angular velocity ω=vR, then what will be the angular momentum of the disc about bottommost point in this new situation?

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The disc starts rotating anticlockwise with the same angular velocity $ω = \frac{v}{R}$ , then what will be the angular momentum of the disc about bottommost point in this new situation?