A large grinding wheel in the shape of a solid cylinder of radius 0.370m is free to rotate on a frictionless, vertical axle. A constant tangential force of 470 N applied to its edge causes the wheel to have an angular acceleration of 0.800 rad/s2. (a) What is the moment of inertia of the wheel?
F*r = I*alpha
where F is the force, r is the radius, I is the moment of inertia, and alpha is the radial acceleration
470*0.370 = I * 0.8
To find the moment of inertia of the wheel, we can use the equation:
τ = I * α
where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.
In this case, the torque exerted by the tangential force is equal to the product of the force and the radius of the wheel. So, the torque (τ) is given by:
τ = F * r
where F is the force applied and r is the radius of the wheel.
Given:
Force (F) = 470 N
Radius (r) = 0.370 m
Angular acceleration (α) = 0.800 rad/s^2
Substituting these values into the torque equation, we have:
470 N * 0.370 m = I * 0.800 rad/s^2
Now, rearranging the equation to solve for the moment of inertia (I):
I = (470 N * 0.370 m) / 0.800 rad/s^2
Calculating this numerical expression will give us the moment of inertia of the wheel.