Question

a uniform disk with mass 37.3 kg and radius 0.290 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 32.0 n is applied tangent to the rim of the disk. what is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.270 revolution?\

Answer 1

The magnitude of the resultant acceleration of a point on the rim of the disk after it has turned through 0.270 revolution is 2.52 m/s².

To find the magnitude of the resultant acceleration of a point on the rim of the disk, we need to use the equation:

a = angular acceleration × radius

First, let's find the angular acceleration. We can use the equation:

α = τ / I

where α is the angular acceleration, τ is the torque, and I is the moment of inertia. Since the disk is uniform, the moment of inertia can be calculated using the formula:

I = (1/2) × m × r²

Substituting the given values, we find I = 1.065 kg·m². Now we can calculate the torque using:

τ = r × F

Substituting the given values, we find τ = 9.28 N·m. Finally, we can calculate the angular acceleration using:

α = τ / I

Substituting the values, we find α = 8.71 rad/s². Now we can find the magnitude of the resultant acceleration using the formula:

a = α × r

Substituting the given values, we find a = 2.52 m/s². Therefore, the magnitude of the resultant acceleration of a point on the rim of the disk after it has turned through 0.270 revolution is 2.52 m/s².