# Angular Momentum Problem

1. Nov 8, 2007

### Destrio

A uniform flat disk of mass M and radius R rotates about a horizontal axis through its center with angular speed ωi.
a) What is its angular momentum
b) A chip of mass m breaks off the edge of the disk at an instant such that the chip rises vertically above the point at which it broke off. How how above the point does it rise before starting to fall?
c) What is the final angular speed of the broken disk.

Li = Iωi
I of wheel = (1/2)MR^2
Li = ωi(1/2)MR^2

d = Vit + (1/2)at^2

i figure this velocity will be equal to sin(theta)*Vtangential (Vτ)

d = Vτi*sin(theta)*t + (1/2)gt^2

is this correct?

thanks

2. Nov 8, 2007

### Staff: Mentor

OK.

You aren't given the time or the tangential speed. And it rises vertically, so what must be its direction?

Find the distance in terms of the given parameters.

3. Nov 8, 2007

### Destrio

Li = Iωi
I of wheel = (1/2)MR^2
Li = ωi(1/2)MR^2

can i assume there is no horizontal component to the flying chip?
if so, wont Vτ = ωiR at some point

do I have to find distance in terms of R?

4. Nov 8, 2007

### Staff: Mentor

I would.
Yes. That's the speed of the chip as it breaks off.

Find it any way you can, but only in terms of the parameters given. That distance only depends on the initial speed, but that speed is in terms of ωi and R.

5. Nov 8, 2007

### Destrio

Li = Iωi
I of wheel = (1/2)MR^2
Li = ωi(1/2)MR^2

Vτ = ωiR

i need to know when Vf = 0

so Vf = Vo + at
Vf = 0
Vo = Vτ = ωiR
-a = g
0 = ωiR - gt
gt = ωiR
t = ωiR/g

d = ωiR(ωiR/g) + (1/2)g(ωiR/g)^2
d = (ωiR)^2/g + (ωiR)^2/2g
d = 2(ωiR)^2/2g + (ωiR)^2/2g
d = 3(ωiR)^2/2g

this this correct?

thanks

6. Nov 8, 2007

OK.
a = -g

7. Nov 8, 2007

### Destrio

d = ωiR(ωiR/g) - (1/2)g(ωiR/g)^2
d = (ωiR)^2/2g

8. Nov 8, 2007

Good.