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 0.978 rad/s^2. What is the moment of inertia of the wheel?

Answer 1

I = 70.86 kg.m²

Step-by-step explanation:

given,

radius of the cylinder = 0.33 m

Tangential force = 210 N

angular acceleration = 0.978 rad/s²

we know,

`\tau = F * r`

and

`\tau =I * \alpha`

computing both the torque equation together

`F * r = I * \alpha`

`I = (F* r)/(\alpha)`

`I = (210* 0.33)/(0.978)`

I = 70.86 kg.m²

Moment of inertia of the wheel is equal to 70.86 Kg.m²