Rotating Thin Disc Homework: Final Angular Speed

In summary, a horizontal thin disc of mass M and radius R rotating about its horizontal axis with angular speed w experiences no change in speed when a chip of mass m breaks off at the edge of the disc. This can be explained by the conservation of angular momentum, as the decrease in moment of inertia causes a corresponding decrease in angular momentum, leaving the speed unaffected. The disc would only experience a change in speed if there was an external force or energy transfer involved in the process.
  • #1
weesiang_loke
33
0

Homework Statement



A horizontal thin disc of mass M and radius R rotates about its horizontal axis through its centre with angular speed w. If a chip of mass m breaks off at the edge of the disc, what is the final angular speed of the disc?

Homework Equations



Initial rotational kinetic energy = 0.5*I*w2
I = 0.5*M*R2

The Attempt at a Solution


I assume that the chip broke off can be considered as some point mass. So the new moment of inertia of the thin disc, Inew = 0.5*M*R2 - m*R2.
so based on the conservation of kinetic energy:
0.5*I*w2 = 0.5*Inew*wnew2 + o.5*m*(R*w)2
then I substituted in the new moment of inertia and simplified.
what i get for wnew is the same as w.
That means the speed is unaffected.
I am not sure whether this is the correct way to do it? and if so why the speed is not affected by the process?


Thanks for any help give.
 
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  • #2
I am not sure why conservation of energy works here but a safer option would have been conservation of angular momentum. If you use that you would get the same result and that can be explained as the disc's angular momentum decreases corresponding to it's moment of inertia. Will find out if conservation of energy here is really applicable.
 
  • #3
Hi aim1732,
but i still don't get it why the speed still remained the same even when some parts fell off.
 
  • #4
That is because the angular momentum decreased but L=Iw---since there is corresponding decrease in I,w happens to remain same.
 
  • #5
ok. i get it now. thanks...
 
  • #6
hi weesiang_loke! :smile:

(have an omega: ω :wink:)
weesiang_loke said:
Hi aim1732,
but i still don't get it why the speed still remained the same even when some parts fell off.

it's because there was no interaction …

it's exactly the same as if the chip was never part of the disc, but was attached to the same axle by a string, and just happened to be rotating at the same angular speed …

when the string is cut, of course it has no effect on the disc! :biggrin:

only if you were told that the chip "pushed off", or that some energy was lost, would there be an interaction and therefore an effect on the disc :wink:
 
  • #7
hi tiny-tim,
i think i fully understand it now. thanks for the explanation. It's really useful.
 

1. What is a rotating thin disc?

A rotating thin disc is a flat, circular object that is spinning around a central axis. It is often used in physics and engineering experiments to study rotational motion and angular momentum.

2. What is the final angular speed?

The final angular speed is the rate at which the rotating thin disc is spinning at the end of a given time interval. It is measured in radians per second and can be calculated by dividing the final angular displacement by the final time.

3. How is the final angular speed calculated?

The final angular speed can be calculated using the formula: final angular speed = final angular displacement / final time. The angular displacement is the change in angle of the disc, while the final time is the duration of the rotation.

4. How does the final angular speed relate to the initial angular speed?

The final angular speed is directly related to the initial angular speed, as it is the result of the initial angular speed and any external torque acting on the disc. The greater the initial angular speed, the faster the disc will rotate, and the higher the final angular speed will be.

5. How is the final angular speed affected by the mass and radius of the disc?

The final angular speed is not affected by the mass of the disc, as long as the mass is evenly distributed. However, the final angular speed is directly proportional to the disc's radius. A disc with a larger radius will have a higher final angular speed compared to a disc with a smaller radius, given the same initial angular speed and external torque.

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