Work done in rotational motion

In summary, the problem involves a turntable of radius 25 cm and rotational inertia 0.0154 kg m^2 spinning freely at 22.0 rpm with a 19.5-g mouse on its outer edge. The mouse walks to the center, and the task is to find the new rotation speed and the work done by the mouse. Using the equations for inertia of a disk and a ring, the solution to part A is 23.7 rpm. For part B, the method is to equate initial and final angular momenta, and since the mouse is at the center in the final state, only the disk's KE is involved. The mistake in using 1/2MR^2
  • #1
thebest100
5
0

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




W = Kf - Ki
Inertia for a disk is MR^2
Intertia for a ring is (1/2)MR^2


The Attempt at a Solution


solution to part A) 23.7 rpm

i have tried the following but they all seem to be wrong (303 mJ, 3.3 mJ, 127 mJ)

the answer to part B must be in mJ

please give me an idea on how to solve part B.
 
Last edited:
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  • #2
I haven’t checked your calculations for part A, so I’m writing the method anyway.

The initial angular momentum comprises of the ang mom of the disk and ang mom of the mouse. The final ang mom is only of the disk, since the mouse is at the centre. Equate initial and final angular momenta.

Initial KE of disk plus mouse < Final KE of same. The increase is because of the work done by the mouse.

Now plug in the numbers.
 
  • #3
Shooting star said:
I haven’t checked your calculations for part A, so I’m writing the method anyway.

The initial angular momentum comprises of the ang mom of the disk and ang mom of the mouse. The final ang mom is only of the disk, since the mouse is at the centre. Equate initial and final angular momenta.

Initial KE of disk plus mouse < Final KE of same. The increase is because of the work done by the mouse.

Now plug in the numbers.

by "Final KE of same" you mean KE of the disk and the mouse

would the KE of the mouse at the center of the disk be 0, so only the KE of disk would be involved in the final right
 
  • #4
thebest100 said:
by "Final KE of same" you mean KE of the disk and the mouse
Right.

would the KE of the mouse at the center of the disk be 0, so only the KE of disk would be involved in the final right
Yes. Just treat the mouse as a point mass (ignore its rotation).
 
  • #5
Thank You Shooting Star and Doc Al. I found my mistake, I was using 1/2MR^2 for the Inertia of the mouse, instead of MR^2.

Thank you
 

Related to Work done in rotational motion

1. What is work done in rotational motion?

Work done in rotational motion is the product of torque and angular displacement. It measures the amount of energy transferred to or from a rotating object.

2. How is work done calculated in rotational motion?

Work done in rotational motion is calculated using the formula W = τθ, where W is work, τ is torque, and θ is angular displacement. This formula is derived from the definition of work as force multiplied by displacement.

3. What are the units of work done in rotational motion?

The units of work done in rotational motion are joules (J) in the SI system. In the imperial system, the unit of work is foot-pounds (ft-lb).

4. What factors affect the amount of work done in rotational motion?

The amount of work done in rotational motion is affected by the magnitude of the torque applied and the angle through which the object is rotated. The direction of the torque also plays a role, as it determines whether the work is positive or negative.

5. Can work done in rotational motion be negative?

Yes, work done in rotational motion can be negative. If the torque and angular displacement are in opposite directions, the work done will be negative, indicating that energy is being transferred out of the system. This can happen, for example, when an object slows down or comes to a stop due to friction.

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