Bowling ball help on small subpart

1. Dec 2, 2003

rdn98

picture is included.

A bowling ball 28 cm in diameter is slid down an alley with which it has a coefficient of sliding friction of µ = 0.52. The ball has an initial velocity of 11 m/s and no rotation. g = 9.81 m/s2. Note: For a sphere Icm = (2/5)mr2.

a) What is the initial deceleration of the ball?
b) What is the initial angular acceleration of the ball?
c) How long does it take before the ball starts to roll without slipping?
d) If it had been moving 15.4 m/s initially, how long would it have taken the ball to start rolling without slipping?
---------------------------
Okay, I've gotten parts a and b, but I'm stuck on C. If I can get C, I can get D easily.

part a) Just did summation of forces in x direction =ma.
u*m*g=m*a
|a|=5.096 m/s

part b) Net torque=I*alpha
so u*m*g*r=(2/5)MR^2*alpha

part c) Okay, this probably is simple to figure out, but I'm not seeing it. I have my initial deceleration and initial angular acceleration of the ball.
Ok, so what do I do to find the time before the ball starts rolling?

File size:
1 KB
Views:
53
2. Dec 2, 2003

Staff: Mentor

What's the condition for rolling without slipping? The translational speed will decrease, and the rotational speed will increase, until that condition is met.

3. Dec 2, 2003

rdn98

Um, the condition has something to do with friction? So the ball starts to roll when its greater than the force of friction against it?

4. Dec 2, 2003

Staff: Mentor

Well, yes, it has to do with friction: if there were no friction, the ball would never start rolling.

I have no idea what your second sentence means.[b(]

Consider this: as long as the surface of the ball keeps sliding against the floor, friction will act to slow it down.

5. Dec 2, 2003

rdn98

I already knew that the speed of the ball surface with respect to the ground is zero.

But I'm still clueless as to how to set up my calculations. Please excuse my questioning. I'm not a physics whiz like you doc al.

6. Dec 2, 2003

NateTG

The force of friction on the ball exerts torque about the center of rotation. This causes the ball to spin in addition to slowing it down.

You should be able to determine the rotational speed of the ball as a function of time.

You should also be able to determine the linear speed of the ball as a function of time.

When the rotational speed of the ball * the radius of the ball = the linear speed, then the surface of the ball is moving at the same speed as the ball, and it stops slipping.

7. Dec 2, 2003

Staff: Mentor

Well... that's only true when it rolls without slipping. But that's the secret. Now express that mathematically. The speed of the center of the ball (translational) is V. The speed of the bottom of the ball, with respect to the center, is &omega;r (going backwards). So then, the speed of the bottom of the ball with respect to the ground is: V - &omega;r. Setting that equal to zero is the condition for rolling without slipping. So... when V = &omega;r is the point where it rolls without slipping.

Does that make sense?

8. Dec 2, 2003

rdn98

Lol. Okay, I knew I had to use that equation, I should have said that earlier. I feel like I have all the pieces in front of me, but just putting it all together is the obstacle.

One thing though, should I use one of those constant angular acceleration equations? Thats what I'm thinking I got to do.

Last edited: Dec 2, 2003
9. Dec 4, 2003

Rockdog

Yup.

Ok. You know that v=vo+at

well, v=w*r
and w=alpha*t

substitute all that into above equation, and solve for time.

Same idea for part d, just use different vo.