A solid disk of mass M and radius R rotates about its central axis with angular velocity ω. Its rotational kinetic energy is which of the following expressions?

A solid disk of mass M and radius R rotates about its central axis with angular velocity ω. Its rotational kinetic energy is which of the following expressions?

• A. (1/4) M R^2 ω^2
• B. (1/2) M R^2 ω^2
• C. (1/2) M ω^2
• D. (1/2) M v^2 with v = R ω

Answer: (A)

Rotational kinetic energy depends on how mass is distributed about the axis of rotation. It uses KE = (1/2) I ω^2, where I is the moment of inertia. For a solid disk about its central axis, I = (1/2) M R^2. Plugging in gives KE = (1/2) × (1/2 M R^2) × ω^2 = (1/4) M R^2 ω^2. That is the correct expression. The other forms would arise from using a different moment of inertia (for a hoop, I = M R^2, which would give (1/2) M R^2 ω^2) or from treating the entire disk as a single mass at the edge (which would yield (1/2) M R^2 ω^2 as well), rather than accounting for the disk’s distributed mass. Using v = R ω directly in 1/2 M v^2 would imply all the mass moves at the edge, which isn’t how a solid disk behaves.

Rotational kinetic energy depends on how mass is distributed about the axis of rotation. It uses KE = (1/2) I ω^2, where I is the moment of inertia.

For a solid disk about its central axis, I = (1/2) M R^2. Plugging in gives KE = (1/2) × (1/2 M R^2) × ω^2 = (1/4) M R^2 ω^2. That is the correct expression.

The other forms would arise from using a different moment of inertia (for a hoop, I = M R^2, which would give (1/2) M R^2 ω^2) or from treating the entire disk as a single mass at the edge (which would yield (1/2) M R^2 ω^2 as well), rather than accounting for the disk’s distributed mass. Using v = R ω directly in 1/2 M v^2 would imply all the mass moves at the edge, which isn’t how a solid disk behaves.