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A cubical block of side a is moving with...

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

A

`(3v)/((4a)`

B

`(3v)/((2a)`

C

`sqrt3v(sqrt2a)`

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
1
As the external impulse acts at 0, we can conserve angular momentum L (initial) can be calculated by the general formula `L_i=I_(CM)omega+MV_(CM)r_(_|_)`where `r_(_|_)`=`_|_` distance of `V_(CM)` fromO.`L_i=I_(CM)(0)+Mva/2``L_f=I_0omega` as O is stationary after impact.`I_Oomega=Mv(a/2)` Here,`I_O=(Ma^2)/6+M(a/sqrt2)^2=(2Ma^2)/3``because (2Ma^2)/3omega=(Mva)/2` `omega=(3v)/(4a)`
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