Question

A cubical block of side 'a' 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

Answer 1

The angular speed of the block after it hits o is  ω=3v / 4a

Explanation:

τ = r x F

Li = Lf

L = r x p = (a/2 J x MVi) = - MVa/2 n

I = Ma^2 / 6

Lf = Iω

Conserving angular momentum about O (just before and just after ) is

mv a/2 = [ma^2 / 6. ma^2 /2] ω

ω=3v / 4a

Thus the angular speed of the block after it hits o is  ω=3v / 4a

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