Question

A cord is wrapped around the rim of a wheel 0.235 m in radius, and a steady pull of 43.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center.

The moment of inertia of the wheel about this shaft is 5.15 kg⋅m2 Compute the angular acceleration of the wheel.

Answer 1

The angular acceleration of the wheel is approximately 8.35 rad/s^2. Solving for α, we find that the angular acceleration of the wheel is approximately 8.35 rad/s2.

Step-by-step explanation:

To calculate the angular acceleration of the wheel, we can use Newton's second law for rotational motion:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

From the given information, we know that the torque is 43.0 N and the moment of inertia is 5.15 kg⋅m2.

Substituting these values into the formula, we get:

43.0 N = 5.15 kg⋅m2 α

Solving for α, we find that the angular acceleration of the wheel is approximately 8.35 rad/s2.