# Homework Help: Bullet Strikes Wheel - find rotations/second

1. Apr 25, 2012

### yesiammanu

1. The problem statement, all variables and given/known data
The wheel shown in the sketch below, is comprised of two 50.0 cm long thin rods of negligible mass and a thin metal ring with a mass of 10.00 kg, mounted on a very low friction bearing. A 100 gram lead weight is shot horizontally at the stationary wheel with an initial speed of 50.0 m/s and sticks to the rim of the wheel. What is the rotational speed of the wheel plus lead mass immediately after the collision? Give your answer in revolutions/second.

2. Relevant equations
I = mr^2
L = Iω
KE = 1/2 mv^2
KErot = 1/2 Iω^2

3. The attempt at a solution
1/2 mv^2 = 1/2 Iω^2
1/2(.1 kg) (50 m/s)^2 = 1/2( 10.1 kg )(.25 m)^2 ω^2
ω^2 = (125 kg * m^2 / s^2)/ .315625 kg * ms^2)
ω = 19.9 / s
Rotations = ω/2pi = 19.9/2pi = 3.16 rev/s

The answer is .315 RPS so I'm not sure what I'm doing wrong

2. Apr 25, 2012

### Staff: Mentor

Energy is not conserved--there's an inelastic collision taking place.

But what is conserved?

3. Apr 25, 2012

### yesiammanu

Is linear momentum/angular momentum conserved? I thought you couldn't go between the two

4. Apr 25, 2012

### Staff: Mentor

One is, one isn't. Which?
What do you mean?

5. Apr 25, 2012

### yesiammanu

I'd assume linear momentum isn't conserved because the momentum absorbed would be redirected to the earth, but doesn't the bullet have no angular momentum? I tried to think of treating it as a tangential force but there's no acceleration

6. Apr 25, 2012

### Staff: Mentor

Good.
Sure it does. L = r X p.

7. Apr 25, 2012

### yesiammanu

Since r is the vector position relative to the origin, it would be .25m (where the origin in the center of the wheel, and the radius is .25m) and the p would be linear momentum, m*v, .1kg * 50 m/s = 5 kg * m/s

It's a tangent so the ABsin(theta) is simply AB, and that would be (5 kg * m/s)(.25m) = 1.25 m^2/s

L= Iw, I is MR^2 for a thin cylindrical shell

1.25 m^2/s = (10.1 kg)(.25 m)^2 w