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 260 n applied to its edge causes the wheel to have an angular acceleration of 0.810 rad/s2. (a) what is the moment of inertia of the wheel?

Question

Lailah

Physics

9 March, 01:07

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 260 n applied to its edge causes the wheel to have an angular acceleration of 0.810 rad/s2. (a) what is the moment of inertia of the wheel?

+1

Answers (1)

Know the Answer?

Kenneth Guerra · Answer 1

Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.

α = 0.81 rad/s², the angular acceleration

r = 0.33 m, the radius of the weel

F = 260 N, the applied tangential force

The applied torque is

T = F*r

= (260 N) * (0.33 m)

= 85.8 N-m

By definition,

T = I*α

Therefore,

I = T/α

= (85.8 N-m) / (0.81 rad/s²)

= 105.93 kg-m²

Answer: 105.93 kg-m²