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Question

A disc of mass 'M' and radius 'R' is rolling with an angular speed of ω rad/s on a horizontal plane as shown in the figure. The magnitude of angular momentum of the disc about the origin O is


A
(12)MR2ω
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B
MR2ω
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C
(32)MR2ω
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D
2MR2ω
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Solution

The correct option is C (32)MR2ω
The angular momentum of a body L may be expressed as the sum of two parts,
(a) one arising from the motion of the centre of mass of the body (Lorbital)
(b) the other from the motion of the body with respect to its centre of mass (Lspin)
i.e., Ltotal=LC.M+rC.M×p
Ltotal=LC.M.+M(C.M×vC.M)
For this problem
LC.M=Iω=12MR2ω and
M(rC.M×vC.M)=MRvCM=MR(Rω)
M(rC.M×vC.M.)=MR2ω
Ltotal=12MR2ω+MR2ω=32MR2ω

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