A uniform disk of radius 1.2 m and mass 0.60 kg is rotating at 25 rad/s around an axis that passes through its center and is perpendicular to the disk. A rod makes contact with the rotating disk with a force of 4.5 N at a point 0.75 m from the axis of rotation as shown. The disk is brought to a stop in 5.0 s. What is the coefficient of kinetic friction for the two materials in contact?

Question

Quizplus · Accepted Answer

First, we need to use the work-energy principle. The work done by the rod is equal to the change in kinetic energy of the disk:
$W = \Delta K
$
$W = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2
$
where $I$ is the moment of inertia of the disk, $\omega_i$ is the initial angular velocity, and $\omega_f$ is the final angular velocity (zero, since the disk comes to a stop). Substituting in the given values, we get:
$4.5 	ext{ N} \cdot 0.75 	ext{ m} = \frac{1}{2} (0.5 m r^2) (0 - (25 	ext{ rad/s})^2)
$
Simplifying and solving for $r$, we get:
$r = 1.2 	ext{ m}
$.
Next, we can use the equation for the frictional torque on a rotating disk:
$	au_f = I \alpha
$
where $	au_f$ is the frictional torque, $I$ is the moment of inertia of the disk, and $\alpha$ is the angular acceleration. Substituting in the given values, we get:
$	au_f = (0.5 m r^2) \left(-\frac{\omega_i}{t}ight)
$
where $t$ is the time it takes for the disk to come to a stop (5.0 s). Solving for $	au_f$, we get:
$	au_f = 0.75 	ext{ N} \cdot 	ext{m}
$.
Finally, we can use the equation for the coefficient of kinetic friction:
$\mu_k = \frac{	au_f}{r F}
$
where $F$ is the force applied by the rod. Substituting in the given values, we get:
$\mu_k = \frac{0.75 	ext{ N} \cdot 	ext{m}}{(1.2 	ext{ m})(4.5 	ext{ N})}
$
Simplifying, we get:
$\mu_k = 0.64
$
Therefore, the coefficient of kinetic friction for the two materials in contact is 0.64, which corresponds to choice C.

A Uniform Disk of Radius 1