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A flywheel of moment of inertia 2 "kg-m"...

A flywheel of moment of inertia `2 "kg-m"^(2)` is rotated at a speed of `30 "rad/s"`. A tangential force at the rim stops the wheel in 15 second. Average torque of the force is

A

4 N-m

B

2 N-m

C

8 N-m

D

1 N-m

Text Solution

Verified by Experts

The correct Answer is:
A
Given, `I=2 "kg-m"^(20),omega=30 "rad s"^(-1) and t=15 s`
Average torque of the force, `tau=Ialpha=I(omega)/(t)=2xx(30)/(15)=4 "N-m"`
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