Answer
(a) At noon, the torque on the clock mechanism is 0.
(b) At 9:00 am, the torque on the clock mechanism is $~794~N\cdot m$
Work Step by Step
(a) The magnitude of the torque can be expressed as $\tau = r~F~sin~\theta$, where $r$ is the displacement between the rotation axis and the point where the force is applied, $F$ is the force, and $\theta$ is the angle between the force vector and the displacement vector $r$.
We can assume that the force of gravity pulls down on the hour hand at the center of gravity, that is, at the midpoint.
We can find the magnitude of the torque at noon:
$\tau = r~ F~sin~\theta$
$\tau = r~ mg~sin~\theta$
$\tau = (1.35~m)(60.0~kg)(9.80~m/s^2)~sin~180^{\circ}$
$\tau = 0$
At noon, the torque on the clock mechanism is 0.
(b) We can find the magnitude of the torque at 9:00 am:
$\tau = r~ F~sin~\theta$
$\tau = r~ mg~sin~\theta$
$\tau = (1.35~m)(60.0~kg)(9.80~m/s^2)~sin~90^{\circ}$
$\tau = 794~N\cdot m$
At 9:00 am, the torque on the clock mechanism is $~794~N\cdot m$
