Question

A uniform disk with mass 36.8 kg and radius 0.280 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.5 N is applied tangent to the rim of the disk. What is the angular acceleration of the disk?

Answer 1

The angular acceleration of the disk is 1.8 rad/s^2.

Explanation:

To find the angular acceleration of the disk, we can use the equation: τ = Iα where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The moment of inertia for a uniform disk rotating about its center is given by:I = (1/2)MR^2 where M is the mass of the disk and R is the radius. Plugging in the values into the equation, we have:τ = (1/2)(36.8 kg)(0.280 m)^2 αSubstituting the given torque of 29.5 N, we can solve for α: 29.5 N = (1/2)(36.8 kg)(0.280 m)^2 α Solving for α, we find that the angular acceleration of the disk is 1.8 rad/s^2.