Arizona State University (ASU) PHY 131 Exam 1 Practice Test

A solid disk of mass 4 kg, radius 0.2 m, rolling without slipping with angular speed ω = 6 rad/s. What is its total kinetic energy?

2.88 J

9.60 J

6.72 J

4.32 J

Rolling without slipping means the disk has both translational and rotational kinetic energy, linked by the velocity v of the center of mass to the angular speed via v = ωR. The total kinetic energy is the sum of translational and rotational parts: KE = 1/2 m v^2 + 1/2 I ω^2. For a solid disk, I = (1/2) m R^2. Using v = ωR, this becomes KE = 1/2 m (ωR)^2 + 1/2 (1/2 m R^2) ω^2 = (3/4) m ω^2 R^2.

Plugging in m = 4 kg, ω = 6 rad/s, R = 0.2 m: R^2 = 0.04 and ω^2 = 36, so KE = (3/4) × 4 × 36 × 0.04 = (3/4) × 5.76 = 4.32 J.

Equivalently, translational KE = 1/2 m v^2 with v = ωR = 1.2 m/s gives 0.5 × 4 × 1.44 = 2.88 J, and rotational KE = 1/2 I ω^2 with I = (1/2) m R^2 = 0.08 kg·m^2 gives 0.5 × 0.08 × 36 = 1.44 J; sum = 4.32 J.