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?

Question

Quizplus · Accepted Answer

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 initial moment of inertia is:$$I_i = 24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$$The final moment of inertia is:$$I_f = 24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } + m r^2$$where $m$ is the mass of the clay and $r$ is the distance from the rotational axis.The initial angular speed is:$$\omega_i = 40.0 \mathrm {~rpm} = 4.19 \mathrm {~rad/s}$$The final angular speed is:$$\omega_f = 32 \mathrm {~rpm} = 3.35 \mathrm {~rad/s}$$Substituting these values into the conservation of angular momentum equation, we get:$$24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } \times 4.19 \mathrm {~rad/s} = (24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } + m (1.2 \mathrm {~m})^2) \times 3.35 \mathrm {~rad/s}$$Solving for $m$, we get:$$m = 4.2 \mathrm {~kg}$$

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