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Question

Physics Question on System of Particles & Rotational Motion

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

A

3v/4a3v/4a

B

3v/2a3v/2a

C

3/2a\sqrt3/\sqrt2a

D

Zero

Answer

3v/4a3v/4a

Explanation

Solution

r2=a22r^2=\frac{a^2}{2}
Net torque about O is zero.
Therefore, angular momentum (L) about O will be conserved,
or Li=LfL_i=L_f
Mv(a2)=I0?=(Icm+Mr2)?Mv\bigg(\frac{a}{2}\bigg)=I_0?=(I_{cm}+Mr^2)?
=\biggl\\{\bigg(\frac{Ma^2}{6}\bigg)+M\bigg(\frac{a^2}{2}\bigg)\biggl\\}?
=23Ma2?=\frac{2}{3}Ma^2?
?=3v4a?=\frac{3v}{4a}