Question
A cubical block of side 'a' is moving with a velocity 'v' on a smooth horizontal plane as shown in the figure. It hits a ridge at point O. The angular speed of the block after it hits O is:
Answer: Option A
:
A
Conserving Angular momentum before and after collision
mv×a2=Iω
Now, I=m(√2a)212+m(a√2)2
=ma2(16+12)
=23ma2.
∴ω=mva2×32ma2=3v4a.
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:
A
Conserving Angular momentum before and after collision
mv×a2=Iω
Now, I=m(√2a)212+m(a√2)2
=ma2(16+12)
=23ma2.
∴ω=mva2×32ma2=3v4a.
Was this answer helpful ?
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