# Pulley and angular momentum (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### candyq27

Hi. I have this problem and I really need help with it. I'm not even sure where to start. Here is the problem:
A horizontal solid disk of mass M and radius R rotates at an angular velocity of w with respect to an axis perpendicular to the disk at its center. Assume that the axis is perfectly frictionless, so that the disk rotates freely.
a. The moment of inertia for a solid disk is .5MR^2. What is the angular momentum of the disk?
From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of R/2 from the axis. The total mass of the sand deposited is M/2.
b. After all the sand is in place, wha is the final angular velocity of the disk? Express your answer in terms of the initial angular velocity w.
c. Calculate the initial and final values of the kinetic energy in the system. Is kinetic energy conserved in this situation? Explain why or why not.

Please help me get started on each step of this problem. I'm not even sure where to begin. Thank you!

#### Doc Al

Mentor
Please give the problems a try and show your work in order to get help. To start with, look up the definitions of angular momentum, moment of inertia, and rotational kinetic energy. For part (b) consider what might be conserved.

Homework Helper
I suggest you start with the definition of angular momentum and the law of its conservation.

Edit: late again.

#### candyq27

Thanks for replying. I'm still not sure even where to begin but here's a try...
I know angular momentum is L=Iw, momentum is p=mv, moment of inertia for the disk is 1/2MR^2, and rotational kinetic energy is K=1/2Iw^2.
So for part (a) I know the moment of inertia is 1/2MR^2 and I need to find out the angular momentum. So since angular momentum is Iw, then...? Is the angular momentum just (1/2MR^2)w?

Now for part (b)...
The initial angular velocity is w. The sand is R/2 and M/2. I need to know the final angular velocity...so I don't know where to go from there. Do I have to use one of the formulas such as w=wo +at (a as in alpha)?

Thanks again.

#### Doc Al

Mentor
So since angular momentum is Iw, then...? Is the angular momentum just (1/2MR^2)w?
You got it.

Now for part (b)...
The initial angular velocity is w. The sand is R/2 and M/2. I need to know the final angular velocity...so I don't know where to go from there.
Big hint: Angular momentum the system (sand + disk) is conserved. But what's the moment of inertia of the system after the sand is added?

#### candyq27

Well since we're adding sand that adds another M/2 and R/2 so would the moment of inertia after the sand is added be 1/2((3/2)Mx(3/2)R)^2?

#### Doc Al

Mentor
No. What's the moment of inertia of the sand by itself? (It's not a disk of sand, but a ring of sand.) The total moment of inertia of the system is just the sum of the parts:
$$I_{total} = I_{disk} + I_{sand}$$

#### candyq27

Ok, so the Itotal= (1/2MR^2) + (MR^2)
but since it tells me that the ring of sand is at M/2 and R/2 that part is (M/2xR/2^2)?
Then I'm not sure where to go from there
So I know the total I and I need to know the final angular w for the disk...do I have to figure out the torques?

#### Doc Al

Mentor
Ok, so the Itotal= (1/2MR^2) + (MR^2)
but since it tells me that the ring of sand is at M/2 and R/2 that part is (M/2xR/2^2)?
Right.

Then I'm not sure where to go from there
So I know the total I and I need to know the final angular w for the disk...do I have to figure out the torques?
Refer to my hint in post #5.

#### candyq27

I don't get it...
I just found the I of the system, but now I have to compare the angular momentums? So the L of the disk was part (a) but now with the sand it's....? ((1/2MR^2)+((M/2)(R/2)^2))w?

#### Doc Al

Mentor
You calculated the angular momentum of the system in part a--that has not changed (the sand dropped onto the disk with zero speed, so it added no angular momentum of its own). Conservation of angular momentum tells you:
$$I_1 \omega_1 = I_2 \omega_2$$

You have the new moment of inertia, now calculate the new angular speed.

#### candyq27

oooh, ok so then i get...
(1/2MR^2)(w) = ((1/2MR^2)+((M/2)(R/2)^2))w2
so w2= [(1/2MR^2)(w)]/[(1/2MR^2)+((M/2)(R/2)^2)]

#### candyq27

So for part (c) I'm not sure what to do either...I know the kinetic energy for rotational motion is KE=1/2 Iw^2
So for the initial kinetic energy it would be 1/2(1/2MR^2)w^2?
and for the final kinetic energy it would be 1/2[(1/2MR^2)+((M/2)(R/2)^2)]w^2?
Then would the kinetic energy not be conserved because of friction?

#### Doc Al

Mentor
Please simplify that expression for $\omega_2$ before my head explodes.

Rotational KE is simply $1/2 I \omega^2$. (Calculate it twice: before and after the sand falls.) The sand hitting the moving disk is an inelastic collision; mechanical energy is "lost" to thermal energy.

Last edited:

#### candyq27

I'm afraid to simplify incase it's wrong but is it just w2=wo/2? no that doesnt look right
for the KE was what i said right for the formulas? thanks so much!

#### candyq27

Ok I simplified the equation for part (b) and i got w2= 4wo/5
That makes more sense now.
For part (c) I got KEi=1/4MR^2w^2 and KEf=5/16MR^2w^2

#### Doc Al

Mentor
Ok I simplified the equation for part (b) and i got w2= 4wo/5
That looks good.
For part (c) I got KEi=1/4MR^2w^2 and KEf=5/16MR^2w^2
Careful how you write these terms. In KEi, w is the original speed; but in KEf, you are using w to mean the final speed--confusing! Rewrite KEf in terms of the original w so that the variables all mean the same thing. (Note that 5/16 > 1/4, so left as is it looks like the system gained energy!)

#### candyq27

Ok so for the final kinetic energy using the original w I get 1/5MR^2w^2. That makes more sense because 1/5 is less than 1/4. Thanks so much!

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving