Question

A constant torque is applied to a rigid body whose moment of inertia is 4.0kg-m2 around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20.0 rad/s in 10.0 s, what is the applied torque?

Answer 1

The applied torque on a rigid body with a moment of inertia of 4.0 kg-m², that starts from rest and reaches an angular velocity of 20.0 rad/s in 10.0 s, is calculated to be 8.0 N-m.

Step-by-step explanation:

To determine the applied torque on a rigid body that starts from rest and attains an angular velocity of 20.0 rad/s in 10.0 s with a moment of inertia of 4.0 kg-m², we use the following

formulas for angular motion:

• Angular acceleration (α) = ∆ω / ∆t
• ∆ω = Final angular velocity (ωf) - Initial angular velocity (ωi)
• Torque (τ) = Moment of inertia (I) × Angular acceleration (α)

Given that the initial angular velocity (ωi) is 0 rad/s (since it starts from rest) and the final angular velocity (ωf) is 20.0 rad/s, we find the angular acceleration as follows:

α = (20.0 rad/s - 0 rad/s) / 10.0 s = 2.0 rad/s²

Now, we can calculate the torque:

τ = 4.0 kg-m² × 2.0 rad/s² = 8.0 N-m

The applied torque is therefore 8.0 Newton-meters (N-m).