A wheel of moment of inertia of 5.00 kg-m² starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s? A) 57.6 J B) 64.0 J C) 91.9 J D) 122.0 J E) 78.8 J

Question

Quizplus · Accepted Answer

The rotational kinetic energy (KE_rot) of the wheel can be calculated using the formula KE_rot = 0.5 * I * ω^2, where I is the moment of inertia and ω is the angular velocity. First, we need to find the angular acceleration (α) using the torque (τ) and moment of inertia (I) with the formula τ = I * α, giving us α = τ / I = 3.00 N-m / 5.00 kg-m^2 = 0.60 rad/s^2. Then, we find the angular velocity (ω) after 8.00 s using ω = α * t = 0.60 rad/s^2 * 8.00 s = 4.80 rad/s. Finally, we calculate the rotational kinetic energy: KE_rot = 0.5 * 5.00 kg-m^2 * (4.80 rad/s)^2 = 57.6 J.

A Wheel of Moment of Inertia of 5