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

Answer:

I = 70.86 kg.m²

Explanation:

given,

radius of the cylinder = 0.33 m

Tangential force = 210 N

angular acceleration = 0.978 rad/s²

we know,

$\tau = F \times r$

and

$\tau =I \times \alpha$

computing both the torque equation together

$F \times r = I \times \alpha$

$I = \dfrac{F\times r}{\alpha}$

$I = \dfrac{210\times 0.33}{0.978}$

       I = 70.86 kg.m²

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