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Question

A cubical block of side a is moving with velocity v on a horizontal smooth plane as shown. It hits a ridge at point O. The angular speed of the block after it hits O is



A
3v4a
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B
3v2a
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C
3v2a
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D
Zero
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Solution

The correct option is D 3v4a
After hitting the ridge at point O, block will rotate immediately with angular velocity ω about axis of rotation passing through O.


τext=0 about point O.
(since torque due to N and mg cancel)
Hence, applying angular momentum conservation:
Li=Lf
Mv(a2)=I0×ω ...(i)


MOI about O from parallel axis theorem,
I0=ICM+Md2
I0=Ma26+M(a2)2
I0=2Ma23
Substituting in Eq. (i),
Mv(a2)=(2Ma23)ω
ω=3v4a

Physics

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