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 fo the block after it hits O is
A
`3V/(4a)`
B
`3V/(2a)`
C
`sqrt(3V) / (sqrt(2a))`
D
zero
Text Solution
Verified by Experts
The correct Answer is:
A
`r = sqrt2(a)/(2)` or `r^2 = (a^2)/(2)` Net torque about O is zero. Therefore, angular momentum (L) about O will be conserved, or `L_i = L_f` `omega = {(Ma^2)/(6) +M((a^2)/(2))}omega = (2)/(3)Ma^2omega = (3v)/(4a)`
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