Record and turntable

gills

1. The problem statement, all variables and given/known data

Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity w_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.



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

(b)Assume that the turntable deccelerated during time [tex]\Delta[/tex]t before reaching the final angular velocity ([tex]\Delta[/tex]t 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,[tex]\tau[/tex] , acting on the bottom disk due to friction with the record?
Express the torque in terms of I_t, w_i, w_f, and [tex]\Delta[/tex]t

2. Relevant equations

K_i - K_f = [tex]\Delta[/tex]K lost to friction --->

(1/2)I_1*w_1i^2 + (1/2)I_2 * w_2i^2 = (1/2)I_1 * w_1f^2 + (1/2)I_2 * w_2f^2



3. The attempt at a solution

Ok, i've only attempted part (a) at the moment so i need to concentrate on that first. then i'll move onto part (b)

What is confusing is what the question is asking for. delta K is just the initial - final energies, but where does that stand in the solution they are looking for.

This is what i know --->
K_i = (1/2)I_t * w_i^2

K_f = (1/2)w_f^2(I_t + I_r)

How and where do I get rid of the w's and where is delta K in the solution?

Doc Al

You don't need delta K, you just need to express K_f in terms of K_i and the other variables. First find w_f. You'd do that just like you did in that similar problem the other day. Once you find w_f, then worry about finding K_f and expressing it the way they want it. (It will take a little algebraic rearranging, that's all.)

glueball

Rotational kinetic energy is not conserved?
But there is a rotational quantity that is conserved. What is it?

gills

I've already figured w_f from the first part of this problem that i didn't post--->

w_f = (I_t * w_i)/(I_t + I_r) ---> i plug this into K_f --->

K_f = (1/2)w_f^2(I_t + I_r) ---> so --->

(1/2)[(I_t * w_i)/(I_t + I_r)]^2 * (I_t + I_r) ---> reduces to --->

(1/2) * (I_t * w_i)^2 / (I_t + I_r) ---> a little stuck here. i'm trying to figure out how i can get a K_i out of this expression...

saket

First thing, why do you need delta K? Neither it has been asked, nor it appears in the two equations you have quoted!
You need K_f in terms of I_t, I_r and K_i. In the above two equations, you have two unknowns, w_f and w_i. To get a required condition, you need just one more equation. So, somehow try to relate w_f and w_i.
Hint: You know some conservation laws, right?

saket

Substitute unwanted w_i in terms of K_i.

gills

(K_i * I_t) / (I_t + I_r) ---> i believe this is correct....Now onto the second part...

saket

ya.. show ur attempt on the second part.
i would just like to point out one thing: note the striking similarity between

a particle, mass m₁, with speed v_i collides and sticks to a second particle, mass m₁, at rest.
conservation of momentum: (m₁+ m₂).v_f = m₁.v_i

a disk, mass moment of inertia I₁, with rotational kinetic energy K_i collides and sticks to a second disk, mass moment of inertia I₂, at rest.
and in this problem: (I₁+ I₂).K_f = I₁.K_i

Have fun.

gills

part (b) solved:

w_f = w_i - [tex]\alpha[/tex]*[tex]\Delta[/tex]t --->

-[tex]\alpha[/tex] = (w_f - w_i) / [tex]\Delta[/tex]t --->

[tex]\tau[/tex] = I*[tex]\alpha[/tex] --->

[tex]\tau[/tex] = I[tex]^{t}[/tex] * (w_f - w_i) / [tex]\Delta[/tex]t

Doc Al

Good! Realize that this is just the angular form of impulse. Just like linear impulse equals change in linear momentum, angular impulse equals change in angular momentum:
[tex]\tau \Delta t = \Delta L = I \Delta \omega[/tex]