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
B
3v2a
C
3v2a
D
Zero

Solution

The correct option is D 3v4aAfter 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(a√2)2 ⇒I0=2Ma23 Substituting in Eq. (i), Mv(a2)=(2Ma23)ω ∴ω=3v4aPhysics

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