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Question

A cubical block is moving on a horizontal smooth plane. It hits a ridge at point O. The angular speed of the block after it hits O is


A
10 rad/s
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B
18 rad/s
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C
12 rad/s
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D
15 rad/s
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Solution

The correct option is D 15 rad/s
Let us suppose angular speed of the block after it hits O is ω about axis passing through O.
From the diagrams shown below:




When the block collides with the ridge, the impulsive force which operates will pass through point O.
Therefore, net torque will be zero about O.
On applying law of conservation of angular momentum.
Li=Lf
mvl2=IOω
where l is the side of the cube.

MOI of cubical block about point O just as it hits the ridge and starts to rotate is given by parallel axis theorem:
IO=ml26+m(l2)2

5×20×12=5×126+5×(12)2ω
50=103ω
ω=15 rad/s
The angular speed of block after it hits O is 15 rad/s.
Hence, option (d) is the correct answer.

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