A light-weight potter's wheel, having a moment of inertia of $24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ is spinning freely at 40.0 rpm. The potter drops a small but dense lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?
A) 4.2 kg
B) 2.8 kg
C) 3.7 kg
D) 4.6 kg

Question

Quizplus · Accepted Answer

The correct answer is A.Explanation:The angular momentum of the system is conserved, so:$$I_i\omega_i = I_f\omega_f$$where $I_i$ and $\omega_i$ are the initial moment of inertia and angular speed, and $I_f$ and $\omega_f$ are the final moment of inertia and angular speed.The moment of inertia of the wheel and clay is:$$I_f = I_i + mr^2$$where $m$ is the mass of the clay and $r$ is the distance from the rotational axis.Substituting into the conservation of angular momentum equation, we get:$$I_i\omega_i = (I_i + mr^2)\omega_f$$Solving for $m$, we get:$$m = \frac{I_i(\omega_i - \omega_f)}{r^2\omega_f}$$Plugging in the given values, we get:$$m = \frac{(24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 })(40.0 \mathrm {~rpm} - 32 \mathrm {~rpm})}{(1.2 \mathrm {~m} )^2(32 \mathrm {~rpm})}$$$$m = 4.2 \mathrm {~kg}$$

A Light-Weight Potter's Wheel, Having a Moment of Inertia Of 24