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Angular Analogue of Linear Momentum

A spherical b...

Question

A spherical body of radius 'R' rolls on a horizontal surface with linear velocity 'v'. Let $L_{1}$ and $L_{2}$ be the magnitudes of angular momenta of the body about centre of mass and point of contact P. Then,

L_{2} = 2 L_{1}

; if radius of gyration K = R

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L_{2} = 2 L_{1}

; for all cases

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L_{2} > 2 L_{1}

; if radius of gyration K < R

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L_{2} > 2 L_{1}

; if radius of gyration K > R

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct option is D $L_{2} > 2 L_{1}$ ; if radius of gyration K > R
$L_{1} = I w = M K^{2} w . . . . . (1)$
$L_{2} = I w + M R v$
$= M K^{2} w + M R (w R)$ (as v = Rw)
$= M w (K^{2} + R^{2}) . . . (2)$
From equations (1) and (2), we can see that
$L_{2} = 2 L_{1}$
when K = R
and $L_{2} > 2 L_{1}$
when K > R

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