Calculating Final Speed & Kinetic Energy Loss for Two Rotating Disks

[Naeem]

Q. A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at finitial = 198 rpm. A second disk of mass M2 = 198 g and radius R2 = 5 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed ffinal.

a) What is ffinal? Please give your answer in units of rpm, but do not enter the units.

Ans. Since there are no external torques on the system, Angular momentum must be conserved.

Finally, the total angular momentum is due to both disks spinning:


Lf = I1w1 + I2w2 = MR2Wf

Since, Li = Lf

1/2 MR2wi = MR2Wf

So, Wf = 1/2Wi

I tried to put Wi as 198 rpm and found out Wf to be 99, which the computer says is wrong.

b) In the process, how much kinetic energy is lost due to friction?

Ans. The initial kinetic energy is 1/2 Iiw12, and the final Kinetic energy is

K2 = 1/2( I1 + 12 )w2

There fore the fraction lost is :

| Delta K | / K1

which is ,

(1/2( I1 + 12 )w2 - 1/2 Iiw12)/ 1/2Iw12, which is 2/3 , which the computer says is wrong.

Can anybody help!

 

[Galileo]

Lf = I1w1 + I2w2 = MR2Wf

Since, Li = Lf

1/2 MR2wi = MR2Wf

So, Wf = 1/2Wi

I tried to put Wi as 198 rpm and found out Wf to be 99, which the computer says is wrong.

The initial angular momentum is:
$$I_1\omega_1+I_2\omega_2$$
the final angular momentum is:
$$(I_1+I_2)\omega_f$$
Find $\omega_f$ from this.

 

[Naeem]

Figured out part a , what equations needed for part b, to work with.

 

[Justin Lazear]

The loss is simply the difference in energy between before and after.

--J

 

[Naeem]

You mean,

1/2 mv2final + 1/2 I w2final = 1/2mv2initial + 1/2 I w2initial.

 

[Justin Lazear]

The point of the question is that energy is not conserved.

What you're saying is $E_i = E_f$, which is saying that energy is conserved. But this is not true, as energy isn't conserved, some is lost to friction. What the question's looking for is

$$E_i = E_f + Q$$

where Q is the loss.

--J