Sliding to Rolling motion problem.

In summary: The rolling...v = v0 + at, v0 = 4.60, a = 5ug/2 = 0.490, v = 4.84 m/sx = x0 + v0t + 1/2 at^2, x0 = 0, v0 = 4.84, x = 1.21 mThe difference is 0.06 m.In summary, The ball travels approximately 1.15 meters while sliding and 1.21 meters while
  • #1
Dorothy Weglend
247
2

Homework Statement


A bowling ball is released without spin. How far does it travel before its motion is pure, rolling motion? No data is given, it asks for an order of magnitude estimate.


Homework Equations


k = mv^2/2
k = Iw^2/2 + mR^2w^2/2
I = 2MR^2/5
torque = Ialpha
torque = fk(R)
Work = fk(d)
V = Rw (w = omega, condition for rolling motion)

The Attempt at a Solution


The ball may slide a bit before it starts to roll at all. It will then roll and slide, then, eventually, only roll. There will be energy lost to friction. At the point the ball begins to roll (which is where V = Rw)

Krolling - Ksliding = -fk(d) or

Iw^2/2 + MR^2w^2/2 - Mv^2/2 = -fk(d)

Kinetic friction exerts a torque on the ball:

Ia = fk(R), and since Icm for a sphere is 2MR^2/5 and w = at

fk = 2MRt/5, substitute this in the above and simplify:

2Rw^2/5 + Rw^2/2 - V^2/2 = -2dt/5

and I think I just took a wrong turn some where :uhh:.

A gentle (or not so gentle) prod in the right direction would be very welcome right now.

Thanks so much,
Dorothy
 
Physics news on Phys.org
  • #2
Try thinking of it this way: The ball starts with pure translational motion. The friction slows the translational motion, and starts the ball rotating. So write equations for the translational motion and for the rotational motion, due to the friction. The ball will slow down (but increase rotational speed) until the speed is just right to stop slipping.

You'll find that the time it takes to begin rolling without slipping depends on how fast it was going when released. You'll just have to take a guess at what a reasonable speed for that would be.
 
  • #3
Doc Al said:
Try thinking of it this way: The ball starts with pure translational motion. The friction slows the translational motion, and starts the ball rotating. So write equations for the translational motion and for the rotational motion, due to the friction. The ball will slow down (but increase rotational speed) until the speed is just right to stop slipping.

You'll find that the time it takes to begin rolling without slipping depends on how fast it was going when released. You'll just have to take a guess at what a reasonable speed for that would be.

I'm still not sure I have this right. When the ball is finally rolling along, I should have

Krolling = Iw^2/2 + mR^2w^2/2, and using I = 2MR^2/5 and w = v/r, I end up with

Krolling = 7Mv^2/10

I can get the acceleration from the torque:

sumof t = Ialpha = umgR

which simplifies to

a = 5ug/2

and I would like to use v^2 = v0^2 -2as to get s, that would be nice.

v0 I can just estimate, but how do I get v (the speed of the rolling CM)?

Dorothy
 
Last edited:
  • #4
Urgh.. Defeat... I just can't figure out how to get v. I can get a bunch of equations, but I don't think I know enough math to solve them, or even know if they can be solved :frown:

The best I can do is this:

using a = 5ug/2, and assuming we know the average velocity, then the time is t = v_avg/a, that should be close to the time it takes to convert to rolling friction, right?

So the distance is x = v_avg(t) = v_avg^2/a = v^2(2/5ug).

I found a figure for average velocity of v = 4.60 m/s from some bowling website, and using 0.100 (waxed wood on snow, from my physics book. I used this because I read that they oil the lanes, and this is the closest I could find)

x = 4.6^2(2/(5(0.100)(9.8))) = 9.4 m.

So, I would say order of magnitude of 10^1 m.

I have to say this all seems like a cheat to me, and since a bowling lane is about 18.3 m, this really seems high.

Is this even on the right track?

Dorothy
 
  • #5
Dorothy Weglend said:
I'm still not sure I have this right. When the ball is finally rolling along, I should have

Krolling = Iw^2/2 + mR^2w^2/2, and using I = 2MR^2/5 and w = v/r, I end up with

Krolling = 7Mv^2/10
This is true, but you won't need it.
I can get the acceleration from the torque:

sumof t = Ialpha = umgR

which simplifies to

a = 5ug/2
OK. But realize that this is the acceleration of the ball's surface with respect to its center. What's the translational acceleration of the ball's center of mass? (Hint: Apply Newton's 2nd law twice--for rotation and for translation.)

As I said earlier, two things are going on: The rotation is speeding up; the translation is slowing down. At some point, the ball's surface speed with respect to the center will just equal the ball's translational speed. That's the point where the motion becomes rolling without slipping. Solve for that point by setting up the equations for rotational and translational speed.
 
  • #6
Doc Al said:
.
OK. But realize that this is the acceleration of the ball's surface with respect to its center. What's the translational acceleration of the ball's center of mass? (Hint: Apply Newton's 2nd law twice--for rotation and for translation.).

But I had to multiply alpha (angular acceleration) by R (radius of ball) to get the linear acceleration (a) in that equation. Doesn't that give me the translational acceleration of the center of mass?

As I said earlier, two things are going on: The rotation is speeding up; the translation is slowing down. At some point, the ball's surface speed with respect to the center will just equal the ball's translational speed. That's the point where the motion becomes rolling without slipping. Solve for that point by setting up the equations for rotational and translational speed.

Maybe I don't understand what you mean by surface speed. I will try this some more, now that I have had some sleep.

Thanks, Doc Al. You are really great.

Dorothy
 
  • #7
It's amazing how dense I can be. Ok, I think I have it, thank you. But I am not that great at estimates and order of magnitude stuff. My estimates are bad, in fact :eek:

Here's what I did:

The sliding translational motion, that would have to be:

f = -ma

and the rotational motion should be:

fR = I(alpha)

where f is force of kinetic friction.

If T is the time it takes to 'get rolling', so to speak, then

fRT = Iw (w is final angular velocity)

The condition for rolling motion is w = vf/R, so

fT = Ivf/R^2

Returning to f = - ma, since a = (vf-v0)/t

fT = -m(vf-v0)

where vf, v0 are final and initial velocities of the center of mass.

Using I = 2mR^2/5 as before, I get this really cool relation:

vf = (5/7) v0

To get the distance, I have an average v of 4.6, m/s, so say it starts out at 5.0 m/s. It seems to me that bowling balls don't slow down that much, and here I think I can use energy:

-fd = mvf^2/2 - mv0^2/2

and using a coefficient of friction of 0.100 and v0 = 5.0 m/s, I get about 6 m for the distance to complete rolling, or about 1/3 the distance of the typical bowling lane.

Well, this is the same order of magnitude as before, 10^1, but I feel much better.

I hope this is right (??).

Thanks again,
Dorothy
 
Last edited:
  • #8
Brilliant!

Just to make it clear--probably not necessary, since you've obviously got it--here's another way to view the condition for rolling without slipping: The instantaneous speed of the point of contact of ball with respect to the floor must be zero. (That's what "not slipping" means, of course.) When you caculate alpha (using Newton's 2nd law for rotation) you are finding the angular speed of the ball about its center; when you mulitply that by R, you are find the linear speed of the ball's surface about its center. In order for the ball not to slip, the speed of the surface with respect to the center must exactly cancel the speed of the center with respect to the floor. Make sense?
 
  • #9
:blushing: Oh Wow... thank you, you made my year.

Yes, that makes sense. I think you are also hinting that I could have found an easier way, using this relationship. I will look for that.

This rolling stuff was very confusing, this problem (and you) helped a lot!

Thank you, thank you, thank you, as always.
Dorothy
 

1. What is the difference between sliding and rolling motion?

Sliding motion refers to the movement of an object where there is no point on the object that remains stationary, while rolling motion is the movement of an object where at least one point on the object remains stationary.

2. How can you calculate the velocity of an object in sliding and rolling motion?

In sliding motion, velocity can be calculated using the equation v = d/t, where v is the velocity, d is the distance traveled, and t is the time taken. In rolling motion, velocity can be calculated using the equation v = rω, where v is the velocity, r is the radius of the rolling object, and ω is the angular velocity.

3. What factors affect the transition from sliding to rolling motion?

The transition from sliding to rolling motion is affected by the surface roughness, friction, and the shape and size of the object. An object with a smooth surface and low friction will transition to rolling motion more easily, while a rough surface and high friction will result in more sliding motion.

4. Can an object experience both sliding and rolling motion at the same time?

Yes, an object can experience both sliding and rolling motion at the same time. This is known as mixed or combined motion and occurs when there is both sliding and rolling friction acting on the object.

5. How does the force of gravity affect the sliding to rolling motion problem?

The force of gravity affects the sliding to rolling motion problem by providing a torque on the object, causing it to roll. If the object is on an incline, the force of gravity will also contribute to the sliding motion of the object.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
700
  • Introductory Physics Homework Help
Replies
28
Views
1K
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
818
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
Back
Top