Rotational dynamics about a fixed axis: A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular velocity of 2.40 rev/s, it stops after turning through 18.2 rev with uniform acceleration. The net torque acting on this sphere as it is slowing down is closest to
A) 0.00593 N ∙ m.
B) 0.0372 N ∙ m.
C) 0.0466 N ∙ m.
D) 0.0620 N ∙ m.
E) 0.149 N ∙ m.

Question

Quizplus · Accepted Answer

To find the net torque, we can use the rotational form of Newton's second law, $\tau = I\alpha$, where $\tau$ is the torque, $I$ is the moment of inertia, and $\alpha$ is the angular acceleration. The moment of inertia $I$ for a solid sphere is $\frac{2}{5}MR^2$, where $M$ is the mass and $R$ is the radius. The angular acceleration $\alpha$ can be found using the equation $\alpha = \frac{\Delta \omega}{\Delta t}$, but since $\Delta t$ is not given directly, we use the kinematic equation $\omega_f^2 = \omega_i^2 + 2\alpha\theta$, where $\omega_f$ is the final angular velocity (0 rad/s, since it stops), $\omega_i$ is the initial angular velocity, and $\theta$ is the angular displacement in radians. Convert revolutions to radians and solve for $\alpha$, then calculate $\tau$ using the moment of inertia and the found $\alpha$.

Rotational Dynamics About a Fixed Axis: a Solid Uniform Sphere