A force of 8 N is applied at a point 0.2 m from the pivot, perpendicular to the lever arm. What is the torque? If the body has moment of inertia I = 0.4 kg·m^2, what angular acceleration does this torque produce?

A force of 8 N is applied at a point 0.2 m from the pivot, perpendicular to the lever arm. What is the torque? If the body has moment of inertia I = 0.4 kg·m^2, what angular acceleration does this torque produce?

• A. τ = 1.6 N·m; α = 4 rad/s^2
• B. τ = 1.6 N·m; α = 2 rad/s^2
• C. τ = 0.8 N·m; α = 4 rad/s^2
• D. τ = 8 N·m; α = 0.5 rad/s^2

Answer: (A)

Torque is the rotational effect of a force about a pivot. When the force is perpendicular to the lever arm, the torque is simply the product of the distance and the force: τ = rF. With r = 0.2 m and F = 8 N, that gives a torque of 1.6 N·m. The angular acceleration that torque produces is found from τ = Iα, where I is the moment of inertia. Using I = 0.4 kg·m^2, the angular acceleration is α = τ/I = 1.6 / 0.4 = 4 rad/s^2. So the torque is 1.6 N·m and the resulting angular acceleration is 4 rad/s^2. If the force weren’t perpendicular, you’d multiply by sinθ, and different r or I would change the resulting α accordingly.

Torque is the rotational effect of a force about a pivot. When the force is perpendicular to the lever arm, the torque is simply the product of the distance and the force: τ = rF. With r = 0.2 m and F = 8 N, that gives a torque of 1.6 N·m.

The angular acceleration that torque produces is found from τ = Iα, where I is the moment of inertia. Using I = 0.4 kg·m^2, the angular acceleration is α = τ/I = 1.6 / 0.4 = 4 rad/s^2.

So the torque is 1.6 N·m and the resulting angular acceleration is 4 rad/s^2. If the force weren’t perpendicular, you’d multiply by sinθ, and different r or I would change the resulting α accordingly.