Question

A grinding wheel with a moment of inertia of 2.0 kg•mº is initially at rest. What angular momentum will the wheel have 10.0 s after a 2.5 N·m torque is applied to it?
a. 25 kg•m/s
b. 4.0 kg•m/s
c. 7.5 kg•m/s
d. 0.25 kg•m/s

Answer 1

Answer:

25 kg m^2/s

Explanation:

The angular acceleration due to the application of torque to the grinding wheel is:

$\tau=I\alpha \rightarrow \alpha=\frac{2.5}{2}=1.25$ (in radian per square second)

So, by the basic equation of the angular acceleration, we can get the angular velocity at t = 10 s as follows:

$\omega=\omega_{o}+\alpha t \rightarrow \omega =0+(1.25)(10) =12.5$ rad/s.

So, the angular momentum at t = 10 s:

$L=I\omega=(2)(12.5) = 25$  kg m^2/s