# Linear and Angular Momentum Problem

1. Nov 28, 2004

### xenogizmo

Hey Everyone,
I have this physics problem, and I'm not really sure how to solve it using linear and angular momentum, I know it can be solved otherwise in different methods, but I'm interested in this one..
The question is:

"A bowling ball is given an initial speed of 5 m/s on an alley such that it slides initially without rolling. The coefficient of friction between teh ball and the alley is 0.30. How far must the ball travel until pure rolling motion?"

Now I know that the friction force causes the change in linear momentum, and the change in angular momentum, so we equate them and solve using the formula of work..
But after I do that it gets really messy, and I had to add a negative sign for no reason to get the right answer..

So I would reeeally appreciate it if someone could write a full explanatory solution showing the steps, thanks a lot!

Xeno

2. Nov 29, 2004

### ehild

Do not forget that he work of friction is negative as the force is opposite to the displacement.

ehild

3. Nov 29, 2004

### arildno

1. The equations of motion:
In the horizontal direction, Newton's second law states for the acceleration of the center of mass:
$$-\mu{M}g=Ma_{C.M}$$
(The force acts in the negative horizontal direction)
The torque about the C.M is:
$$-R\vec{k}\times(-\mu{M}g\vec{i})=\frac{2}{5}MR^{2}\dot{\omega}\vec{j}$$
Where R is the radius of the sphere (it will cancel out in your problem later on).
These equations yields the expressions for the C.M velocity and the angular velocity in the period up to pure rolling.

To determine the time T when the ball starts pure rolling, we use the requirement:
$$v_{G}(T)+\omega(T)\vec{j}\times(-R\vec{k})=\vec{0}$$

Hope that helped..

Last edited: Nov 29, 2004
4. Nov 29, 2004

### Cyrus

Could you explain the left side of your equation please. $$-\mu{M}g=Ma_{g}$$
You have equal and opposite forces. But I dont see why you chose -mu*M*g. Also what is A_g on the right side. Is that the acceleration due to gravity, or the center of mass?

Also in the second equation, is R the radius?

5. Nov 29, 2004

### arildno

good catch, cyrus:
I've done some editing; there is only one horizontal force acting on the system.

6. Dec 1, 2004

### Cyrus

If you have time arildno, can you step through the calculations. I tried to do some work on it myself but did not get very far. It seems that you equated the torque equal to = I*omega. But I thought torque is equal to I domega/dt. So what happened to the dt at the bottom. Also where does the R cancel out? That equation leaves you with one R. Thanks for your help!

7. Dec 1, 2004

### arildno

First, I wrote $$\dot{\omega}$$ rather than $$\omega$$

The "dot"-notation is a hallowed tradition reaching back to Newton for writing time-derivatives..
So, we're in agreement here!
I'll post a detailed solution a bit later.

8. Dec 1, 2004

### arildno

For the solution:
We find that the velocity of the C.M must be:
$$v_{G}(T)=v_{0}-\mu{g}t, v_{0}=5m/s$$ (in the $$\vec{i}$$-direction)
The equation for the angular velocity as a function of time is:
$$\omega(t)=\frac{5\mu{gt}}{2R}$$

Now, to determine "T" in the rolling condition, we note that:
$$\omega(T)\vec{j}\times(\vec{-R}\vec{k})=-\frac{5\mu{gT}}{2}\vec{i}$$
Hence, we must have, in the $$\vec{i}$$ direction:
$$v_{0}-\mu{g}T-\frac{5\mu{g}T}{2}=0$$
Or:
$$T=\frac{2v_{0}}{7\mu{g}}$$

This T-value is then used in the distance formula:
$$r(T)=v_{0}T-\frac{\mu{g}T^{2}}{2}=\frac{12v_{0}^{2}}{49\mu{g}}$$

Last edited: Dec 1, 2004