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 conservation of angular momentum principle applies here, where the initial angular momentum of the system (wheel) is equal to the final angular momentum of the system (wheel + clay). The formula for angular momentum is $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.Initial angular momentum = $I_{wheel}\omega_{initial}$Final angular momentum = $(I_{wheel} + I_{clay})\omega_{final}$Given:- $I_{wheel} = 24 \, \mathrm{kg \cdot m^2}$- $\omega_{initial} = 40 \, \mathrm{rpm} = \frac{40}{60} \, \mathrm{rps} \times 2\pi = \frac{40}{30}\pi \, \mathrm{rad/s}$- $\omega_{final} = 32 \, \mathrm{rpm} = \frac{32}{60} \, \mathrm{rps} \times 2\pi = \frac{32}{30}\pi \, \mathrm{rad/s}$- $r = 1.2 \, \mathrm{m}$ (distance from the rotational axis)The moment of inertia of the clay ($I_{clay}$) can be represented as $I_{clay} = mr^2$, where $m$ is the mass of the clay.By equating the initial and final angular momentum and solving for $m$, we get:$$24 \cdot \frac{40}{30}\pi = (24 + m \cdot 1.2^2) \cdot \frac{32}{30}\pi$$Solving for $m$ gives us the mass of the clay as approximately 4.2 kg.

A Light-Weight Potter's Wheel,having a Moment of Inertia Of 24 kg⋅m224 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }24 kg⋅m2

A Light-Weight Potter's Wheel,having a Moment of Inertia Of $24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$