Conservation of angular momentum of a turntable

In summary, the final angular velocity of the two disks is omega_f = (I_t+I_r)*(I_t*2*K_i)/(2*(I_t^2+2*I_t*I_r+I_r^2)).
  • #1
~angel~
150
0
Please help.

Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity omega_i around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I_r. The initial angular velocity of the second disk is zero.

There is friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular velocity.

What is the final angular velocity, omega_f, of the two disks?
Express omega_f in terms of I_t, I_r, and omega_i.

Because of friction, kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final kinetic energy, K_f, of the two spinning disks?
Express the final kinetic energy in terms of I_t, I_r, and the initial kinetic energy K_i of the two-disk system. No angular velocities should appear in your answer.

Assume that the turntable deccelerated during time deltat before reaching the final angular velocity ( deltat is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, \avg{\tau}, acting on the bottom disk due to friction with the record?
Express the torque in terms of I_t, omega_i, omega_f, and deltat.

Ok, I know the answer to the first question, which is (I_t*omega_i)/(I_t+I_r). But I am lost on the whole KE thing. My answer was ((I_t+I_r)*(I_t*K_i))/(2*(I_t^2+2*I_t*I_r+I_r^2)), which was incorrect.

Any help would be appreciated for the last question as well.

Thanks in advance.
 
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  • #2
~angel~ said:
Because of friction, kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final kinetic energy, K_f, of the two spinning disks?
Express the final kinetic energy in terms of I_t, I_r, and the initial kinetic energy K_i of the two-disk system. No angular velocities should appear in your answer.

I'm not sure how you came up with that answer. But you should already know that
[tex]K_i = \frac{1}{2}I_t\omega _i^2[/tex]

So express [tex]\omega _i[/tex] in terms of K_i and I_t and substitute it in the equation for K_f:

[tex]K_f = \frac{1}{2}(I_t + I_r)(\frac{I_t\omega _i}{I_t + I_r})^2[/tex]

So show us whether your answer will be the same as the one in the textbook.
 
  • #3
I worked it out...the reason i got it wrong is because I put Ki instead of 2ki.

The answer is (I_t+I_r)*(I_t*2*K_i))/(2*(I_t^2+2*I_t*I_r+I_r^2)

The answer is long because you need to eliminate any angular velocity variables.
 
Last edited:
  • #4
You mean:


[tex]\frac{(I_t + I_r)(2I_tK_i)}{2I_t^2 + 2I_tI_r + I_r^2}[/tex]

?

Then it is simplified to

[tex]\frac{I_t}{I_t + I_r}K_i[/tex]
 
  • #5
Yeah. Just didnt simplify though. How would you determine the torque when the turntable decelerates?
 
  • #6
Easy,

[tex]\tau_{av} = \frac{\Delta L}{\Delta t}[/tex],

where L is the angular momentum.
 
  • #7
lol...ok. These questions are weird. I found the hardest question to be the easiest for me, and the easy ones to be hard.
Thanks
 
  • #8
heh i got the first 2 easy, but I am totally lost on the third, how do u work out the angular momentum?
 
  • #9
[tex]L = I \omega[/tex]
 
  • #10
heh i got it, but i knew that, i just didnt realize i needed to use torque= I*alpha, nstead of momentum/time. then u solve alpha down into terms of intital and final angular velocity over time, and bingo
 

Related to Conservation of angular momentum of a turntable

1. What is conservation of angular momentum?

Conservation of angular momentum is a physical law that states that the total angular momentum of a system remains constant unless acted upon by an external torque.

2. How does angular momentum apply to a turntable?

When a turntable is rotating, it has angular momentum. This momentum is conserved and will remain constant unless there is an external force or torque acting on the turntable.

3. How is angular momentum conserved in a turntable?

Angular momentum is conserved in a turntable because the spinning motion of the turntable creates a force that counteracts any external torque acting on it. This allows the turntable to maintain its constant rotation.

4. Can the conservation of angular momentum be demonstrated on a turntable?

Yes, the conservation of angular momentum can be easily demonstrated on a turntable. This can be shown by placing an object on the turntable and spinning it. The object will remain in its original position and will not move outward even as the turntable continues to rotate.

5. How is the conservation of angular momentum important in turntable design and operation?

The conservation of angular momentum is important in turntable design and operation because it allows the turntable to maintain a consistent rotation and prevents it from slowing down or stopping due to external forces. This is crucial for accurate playback of music on a turntable.

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