# Mouse onto the edge of phonograph turntable

• tonit
In summary, if the angular velocity of the turntable doesn't change, then the work done by the mouse to walk into the center post is the integral of force "dot" displacement.
tonit

## Homework Statement

$m$mouse $= 0.06kg$

$r = 0.3m$

$\omega = 0.05 rad / s$

Suppose angular speed does not change.
What is the work needed for the mouse to go to the center.

## Homework Equations

$I$$o$$\omega$$o$ = $I$$f$$\omega$$f$

## The Attempt at a Solution

If $\omega$ doesn't change ,then how can I use the above equation.

Since $I$ of mouse would be equal to $mr^2$ then $\omega$ would change. I am stuck here. Oh, and it's not a homework, it's a problem that I was stuck at some time ago. Thanks for any help

tonit said:

## Homework Equations

$I$$o$$\omega$$o$ = $I$$f$$\omega$$f$

## The Attempt at a Solution

If $\omega$ doesn't change ,then how can I use the above equation.
You can't. If ω is fixed then angular momentum is not conserved.

So in this case would the work be equal to: $mr^2$ ?

tonit said:
So in this case would the work be equal to: $mr^2$ ?
That's an expression for rotational inertia, not work.

Can you please state the full problem as it was given?

A 60 gm mouse falls onto the outer edge of a phonograph turntable of radius 30 cm rotating at 33rev/min. How much work must it do to walk into the center post? Assume that the angular velocity of the turntable doesn't change.

Seems like you just need to find out the centripetal force that the mouse feels and then you'll have the work. Though I could be wrong.

hi tonit!
tonit said:
A 60 gm mouse falls onto the outer edge of a phonograph turntable of radius 30 cm rotating at 33rev/min. How much work must it do to walk into the center post? Assume that the angular velocity of the turntable doesn't change.

yes, work done is the integral of force "dot" displacement

so, first, what is the force needed, as a function of r ?

There is a centripetal acceleration associated with rotation at constant angular velocity, $a_c = \omega^2 r$. So if my suspicions are correct, you can interpret that as a force that must be overcome, $F_c = m \omega^2 r$

thanks to all of you. now it is all clear :D

## 1. What is a phonograph turntable?

A phonograph turntable is a device used to play vinyl records. It typically consists of a rotating platter, a tonearm, and a stylus (or needle) that reads the grooves on the record to produce sound.

## 2. How does a phonograph turntable work?

A phonograph turntable works by spinning the record at a constant speed while the tonearm moves across the surface of the record, following the grooves and producing sound vibrations. The stylus then converts these vibrations into electrical signals that are amplified and played through speakers.

## 3. What is the purpose of placing a mouse onto the edge of a phonograph turntable?

Placing a mouse onto the edge of a phonograph turntable is typically done for entertainment or novelty purposes. It allows the mouse to ride along with the spinning record, creating a unique and amusing visual effect.

## 4. Is it safe to place a mouse onto the edge of a phonograph turntable?

In most cases, it is not recommended to place a live mouse onto the edge of a phonograph turntable. The spinning motion and potential loud noise can be distressing and potentially harmful to the mouse. It is best to use a toy or fake mouse for this purpose instead.

## 5. Can placing a mouse onto the edge of a phonograph turntable damage the turntable?

In general, placing a lightweight object such as a toy mouse onto the edge of a phonograph turntable should not cause any damage. However, if the object is too heavy or placed too close to the center of the turntable, it could potentially throw off the balance and cause damage to the turntable's motor. It is always best to use caution and common sense when attempting to do this type of activity.

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