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

Options

A. `(3v)/((4a)`
B. `(3v)/((2a)`
C. `sqrt3v(sqrt2a)`
D. Zero

Answer

1

Explanation

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)`