In the figure, the rotational inertia of the wheel and axle about the center is 12.0 kg · m², the constant force F is 39.2 N, and the radius r is 0.800 m. The wheel starts from rest. When the force has acted through 2.00 m, the rotational velocity $\omega$ acquired by the wheel due to this force will be A) 1.26 rad/s B) 3.33 rad/s C) 3.61 rad/s D) 6.24 rad/s E) 10.3 rad/s

Question

Quizplus · Accepted Answer

The rotational kinetic energy of the wheel due to the force F can be found using the formula:
K_rot = (1/2)Iω^2
where I is the rotational inertia, ω is the angular velocity.

In this case, the work done by the force F in rotating the wheel through 2.00 m can be found using the formula:
W = Fd = 39.2 N x 2.00 m = 78.4 J

This work done is equal to the rotational kinetic energy gained by the wheel due to the force F, so we have:
K_rot = 78.4 J

Substituting the given values of I and solving for ω, we get:
ω = √(2K_rot/I) = 3.61 rad/s

Therefore, the correct answer is choice C, 3.61 rad/s.

In the Figure, the Rotational Inertia of the Wheel and Axle