Question

A turntable with a moment of inertia of 0.011 kg∗m2 rotates freely at 3.1 rad/s. A circular disk of mass 400 g and diameter 22 cm, and initially not rotating, slips down a spindle and lands on the turntable.

a) Find the new angular speed (rad/s)

b) what is the change in kinetic energy?

Answer 1

Answer on Question 56960, Physics, Mechanics, Relativity

Question:

A turntable with a moment of inertia of $0.011kg \cdot m^2$ rotates freely at $3.1\,rad/s$. A circular disk of mass $400g$ and diameter of $22cm$, and initially not rotating, slips down a spindle and lands on the turntable.

a) Find the new angular speed $(rad/s)$.

b) What is the change in kinetic energy?

Solution:

a) We can find the new angular speed of the turntable from the law of conservation of angular momentum:

$$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega,$$

here, $I_1 \omega_1$ is the initial angular momentum of the turntable, $I_2 \omega_2$ is the initial angular momentum of the circular disk (since the circular disk initially not rotating, $I_2 \omega_2 = 0 \frac{kg \cdot m^2}{s}$), $I_1, I_2$ are the moments of inertia of the turntable and circular disk, respectively, $\omega_1, \omega_2$ are the angular speeds of the turntable and circular disk, respectively and $\omega$ is the new angular speed.

By the definition of the moment of inertia of the circular disk we have:

$$I_2 = \frac{1}{2} m_2 r^2 = \frac{1}{2} \cdot 0.4 kg \cdot (0.11 m)^2 = 2.42 \cdot 10^{-3} kg \cdot m^2.$$

Then, we can find the new angular speed of the turntable:

$$\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{(I_1 + I_2)} = \frac{0.011 kg \cdot m^2 \cdot 3.1 \frac{rad}{s} + 2.42 \cdot 10^{-3} kg \cdot m^2 \cdot 0 \frac{rad}{s}}{0.011 kg \cdot m^2 + 2.42 \cdot 10^{-3} kg \cdot m^2} = \frac{0.0341 kg \cdot m^2 \cdot \frac{rad}{s}}{0.01342 kg \cdot m^2} = 2.54 \frac{rad}{s}.$$

b) Let's write the initial kinetic energy of the turntable and the circular disk:

$$KE_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} \cdot 0.011 kg \cdot m^2 \cdot \left(3.1 \frac{rad}{s}\right)^2 = 0.053 J.$$ $$KE_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} \cdot 2.42 \cdot 10^{-3} kg \cdot m^2 \cdot \left(0 \frac{rad}{s}\right)^2 = 0J.$$

The final kinetic energy after the circular disk lands on the turntable will be:

$$\begin{aligned} KE_{final} = \frac{1}{2} I_1 \omega^2 + \frac{1}{2} I_2 \omega^2 = \frac{1}{2} (I_1 + I_2) \omega^2 = \frac{1}{2} (I_1 + I_2) \cdot \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{(I_1 + I_2)^2} \\ = \frac{1}{2} \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{(I_1 + I_2)} = \frac{(0.011 kg \cdot m^2 \cdot 3.1 \frac{rad}{s})^2}{2 \cdot (0.011 kg \cdot m^2 + 2.42 \cdot 10^{-3} kg \cdot m^2)} = \\ = 0.043J. \end{aligned}$$

Then, the change in kinetic energy will be:

$$\Delta KE = KE_{final} - (KE_1 + KE_2) = 0.043J - 0.053J = -0.01J.$$

Sign minus means that we have the loss in kinetic energy.

**Answer:**

a) $\omega = 2.54 \frac{rad}{s}$.

b) $\Delta KE = 0.01J$.

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