A wheel of moment of inertia of 300 kg.m² initially at rest is accelerated by a constant torque of 5.00 N.m over 9.00 s. The wheel's rotational kinetic energy at the end of the 9.00 s is A) 0.150 J. B) 0.296 J. C) 1.00 J. D) 3.38 J. E) 6.67 J.

Question

Quizplus · Accepted Answer

The rotational kinetic energy (KE_rot) can be calculated using the formula KE_rot = 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular velocity. To find ω, we use the formula ω = τ/I * t, where τ is the torque, I is the moment of inertia, and t is the time. Substituting the given values, ω = 5.00 N.m / 300 kg.m^2 * 9.00 s = 0.15 rad/s. Substituting ω back into the kinetic energy formula gives KE_rot = 1/2 * 300 kg.m^2 * (0.15 rad/s)^2 = 3.375 J, which rounds to 3.38 J.

A Wheel of Moment of Inertia of 300 Kg