# Bowling ball problem

1. Dec 9, 2003

### zekester

A bowling ball is thrown such that at the instant it touches the floore it is moving horizontally with a speed of 8m/s and is not rotating. it slides for a time and distance before it begins to roll without slipping. The coefficient of friction between the ball and the floor is .06. What is the final speed of the ball?

I know how to find the horizontal acceleration but I don't know where to go from there.

Any hints?

2. Dec 9, 2003

3. Dec 9, 2003

### nbo10

There's an hard way and an easy way.
The easy way is to use conservation of angular momentum.

JMD

4. Dec 9, 2003

### Staff: Mentor

Angular momentum is not conserved. (The friction exerts a torque.)

5. Dec 9, 2003

### nbo10

Angular momentum is conserved about a point, you have to find the correct point.
and if your looking for the FINAL speed, it's going to be zero.

JMD

Last edited: Dec 9, 2003
6. Dec 9, 2003

### HallsofIvy

Aren't those two statements contradictory? If the final speed is 0, then the angular momentum is 0 which is not true while it's rolling.

Actually once it gets to pure roling without sliding, the friction is no longer a factor (that's the whole point of the wheel!).

7. Dec 9, 2003

### formulajoe

ive got a few formulas here:
alpha = (5 * coeff friction * g)/2R
omega = (5 * coeff friction * g * t)/2R

the 8 m/s is beginning translation motion right? but i cant figure out where to begin. ive gone through a few sheets of paper, and i dont think he gave us enough to start with.

8. Dec 9, 2003

### formulajoe

okay, using 8 = R*omega, i was able to find a time. i did 8 = R (5 * coeff friction *g * t)/2r. i ended up with a time of 5.4 seconds. but im stuck now.

Last edited: Dec 9, 2003
9. Dec 9, 2003

### formulajoe

heres what i did:
alpha = (5 * mu * g) 2*R
alpha = a/r, so a/r = (5 * mu * g)/ 2*R. with this i ended up with an a of 1.47.
i than used this 1.47 in omega = initial omega(which is zero because it was not rotating) + alpha * time.
i found time using the formula:
t = 2 * initial velocity/7 * mu * g, which equals 3.89 s.

than i went with v/r = 0 + 1.47 * t/R
the r's cancel and i ended up with a final velocity of 5.7 m/s.
have i done everything correct?

10. Dec 10, 2003

### nbo10

correct.. You can also use conservation of Angular momentum, about the point where the ball makes contact with the floor.

When the ball first starts
$l=mrv_0$
when the ball starts to roll without sliping
$l=I \omega$
$I = \frac{2}{5}mr^2 + mr^2$
for rolling without slipping you can find
$\omega = \frac{v}{r}$

and you can solve these and find
$v = \frac{5}{7}v_0$
Which is 5.7 m/s

I might have to edit this to get the LaTex to work.

JMD

Last edited: Dec 10, 2003
11. Dec 10, 2003

### nbo10

They are contradictory. I've been ill with the flu and still not thinking clearly all the time. The final speed won't be zero.

JMD

12. Dec 10, 2003