Question

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 210 N applied to its edge causes the wheel to have an angular acceleration of 0.932 rad/s2. What is the moment of inertia of the wheel? (Pick the answer closest to the true value.)A. 27.3 kg m2B. 42.4 kg m2C. 54.9 kg m2D. 74.4 kg m2E. 98.5 kg m2

Answer 1

Answer:

Moment of inertia will be $I=74.356kgm^2$

So option (d) will be the correct answer

Explanation:

We have given radius of solid cylinder r = 0.330 m

Constant tangential force F = 210 N

Angular acceleration $\alpha =0.932rad/sec^2$

We know that torque $\tau =Fr=210\times 0.330=69.3Nm$

We also know that torque is given by $\tau =I\alpha$

So $69.3=I\times 0.932$

$I=74.356kgm^2$

So option (d) will be the correct answer