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Q.

A disc of mass M and radius R is rolling with angular speed $ω$ on a horizontal plane as shown in Figure. The ratio of magnitude of angular momentum of the disc about centre of disc and about the origin O is

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a

1:3

b

1:1

c

2:3

d

2:1

answer is C.

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Detailed Solution

$We know L_{p} = L_{CM} + r \times P_{cm}$

$About CM, L_{1} = Iω = \frac{{mr}^{2}}{2} ω$

$About 'O', L_{1} = Iω + mvr$

$= \frac{{mr}^{2}}{2} ω + m (rω) r = \frac{3}{2} {mr}^{2} ω$

Therefore, ratio is 1/2 : 3/2 = 1:3

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