What is the new rotation speed and work done by a mouse on a spinning turntable?

[Jar9284]

Homework Statement

A turntable of radius 25 cm and rotational inertia 0.0154 kg * m$$^{2}$$ is spinning freely at 22.0 rpm about its central axis, with a 19.5-g mouse on its outer edge. The mouse walks from the edge to the center. Find (a) the new rotation speed and (b) the work done by the mouse.

Homework Equations

For part A

$$I$$(total) = $$I$$(table) + $$I$$(mouse)

$$I$$(initial) = 0.0154 kg * m$$^{2}$$ + MR$$^{2}$$
$$I$$(initial) = 0.0154 kg * m$$^{2}$$ + (.0195)(.25)$$^{2}$$ = .016619 kg * m$$^{2}$$ @ $$\omega$$0 = 22 rpm

$$I$$(now) = 0.0154 kg * m$$^{2}$$ + 0 = 0.0154 kg * m$$^{2}$$ @ $$\omega$$1 = ?

Using conservation of angular momentum

(.016619 kg * m$$^{2}$$) * (22 rpm) = (0.0154 kg * m$$^{2}$$) * $$\omega$$1

$$\omega$$1 = 23.74 rpm <- I know that's right

Part B

I'm assuming that it will be this

W = $$\frac{1}{2}$$$$I$$$$\omega$$$$^{2}_{f}$$ - $$\frac{1}{2}$$$$I$$$$\omega$$$$^{2}_{i}$$

With $$I$$ being the mouse's intertia.

The Attempt at a Solution

See above. I just want to see that the equation for part B would be the right approach for it.

 

[alphysicist]

Hi Jar9284,

Homework Statement

A turntable of radius 25 cm and rotational inertia 0.0154 kg * m$$^{2}$$ is spinning freely at 22.0 rpm about its central axis, with a 19.5-g mouse on its outer edge. The mouse walks from the edge to the center. Find (a) the new rotation speed and (b) the work done by the mouse.

Homework Equations

For part A

$$I$$(total) = $$I$$(table) + $$I$$(mouse)

$$I$$(initial) = 0.0154 kg * m$$^{2}$$ + MR$$^{2}$$
$$I$$(initial) = 0.0154 kg * m$$^{2}$$ + (.0195)(.25)$$^{2}$$ = .016619 kg * m$$^{2}$$ @ $$\omega$$0 = 22 rpm

$$I$$(now) = 0.0154 kg * m$$^{2}$$ + 0 = 0.0154 kg * m$$^{2}$$ @ $$\omega$$1 = ?

Using conservation of angular momentum

(.016619 kg * m$$^{2}$$) * (22 rpm) = (0.0154 kg * m$$^{2}$$) * $$\omega$$1

$$\omega$$1 = 23.74 rpm <- I know that's right

Part B

I'm assuming that it will be this

W = $$\frac{1}{2}$$$$I$$$$\omega$$$$^{2}_{f}$$ - $$\frac{1}{2}$$$$I$$$$\omega$$$$^{2}_{i}$$

With $$I$$ being the mouse's intertia.

I believe you have to account for the turntable's moment of inertia also.

 

[baba89]

Your approach for part A and B is correct. For part B, you can also use the formula W = \Delta KE = \frac{1}{2}I(\omega_f^2 - \omega_i^2) to calculate the work done by the mouse. Both approaches will give you the same result.