Bowling Ball Problem: Is Friction Force All Absorbed in Rotation?

[jbriggs444]

So, let's say it was sliding only (like a block instead of a ball), the work done by sliding friction would = friction force * distance travelled.
But if part of that work is involved in ball rotation acceleration, surely it detracts from the overall ?

Overall what, exactly?

Are you comparing apples with apples? Is the distance traveled in the sliding case (before the block comes to rest) the same as the distance traveled in the skidding/rolling case (before the ball is rotating at the right rate for its reduced velocity)?

 

[rcgldr]

Is the distance traveled in the sliding case (before the block comes to rest) the same as the distance traveled in the skidding/rolling case (before the ball is rotating at the right rate for its reduced velocity)?

No, the rolling ball will slide further because it's rate of linear deceleration is less than that of the sliding box, so it will cover more distance while sliding (higher average velocity since the final velocity is not zero).

haruspex corrected this in his next post. - The rolling ball will slide less distance: the rate of deceleration is the same, but the bowling ball stops sliding as soon as it transitions into rolling (still moving forward, but without sliding), while the box slides until it comes to a complete stop.

These type of problems usually ignore issues like rolling friction, so once the ball is rolling, it doesn't slow down.

 

[haruspex]

the rolling ball will slide further because it's rate of linear deceleration is less than that of the sliding box,

Umm... no.
While sliding, the frictional force is the same for both, so the linear deceleration is the same.
The differences are that:
1. The frictional force acts through a shorter distance in terms of surface contact, so less work is done against friction.
2. While sliding, that work difference goes into supplying the rotational energy. This happens via the torque that the frictional force applies.
3. When the rotation rate matches the linear movement to achieve rolling, the frictional force disappears. This happens over a shorter distance, measured along the ground, than in the sliding box case.
Writing u and v for initial speed and linear speed at rolling, t for the time and s for the horizontal distance to achieve rolling:
$v = u\frac{mr^2}{I+mr^2}$
$t = \frac{u}{\mu g}\frac{I}{I+mr^2}$
$s = \frac{u^2}{2 \mu g}\frac{I(I+2mr^2)}{(I+mr^2)^2}$
The corresponding expressions for a sliding box to stop can be got by letting I tend to infinity.
It is clear that this gives a longer time and a greater distance.

 

[sophiecentaur]

No, the rolling ball will slide further because it's rate of linear deceleration is less than that of the sliding box, so it will cover more distance while sliding (higher average velocity since the final velocity is not zero).

haruspex corrected this in his next post. - The rolling ball will slide less distance: the rate of deceleration is the same, but the bowling ball stops sliding as soon as it transitions into rolling (still moving forward, but without sliding), while the box slides until it comes to a complete stop.

These type of problems usually ignore issues like rolling friction, so once the ball is rolling, it doesn't slow down.

Yes. It is always better to analyse, thoroughly, the problem as originally stated (i.e. with the extras all left out) and, when all the juice has been squeezed out of that lemon, start introducing other factors. This thread suffered from extra considerations being added in far too early. haruspex summed it all up nicely.

 

[dean barry]

Thanks for your time all.
Here a question:
g = 10 m/s/s
A 10 kg bowling ball with a radius of 0.1 m and a sliding friction co-efficient of 0.1 is released at a linear 10 m/s onto the lane (no rotation at this point).
Calculate the distance from the point of contact to the point where synchronous speed is realized.
(and frictional force then = 0)
Calculate the time for the same.
Friction force = 10 * 10 * 0.1 = 10 N
It is true to say that :
Over the distance described above :
Energy lost to friction = loss of linear KE + Gain in rotational KE

 

[haruspex]

the work done by sliding friction would = friction force * distance travelled

I'd just like to emphasise one thing you might not have noticed in my post #38.
The work done by the friction is the force multiplied by the distance that the surfaces have moved in relation to each other. Since the ball starts to rotate immediately, this is somewhat less than the linear distance the ball skids.
Specifically, it is $\hat s = \frac{u^2}{2 \mu g}\frac{I}{(I+mr^2)}$ instead of $s = \frac{u^2}{2 \mu g}\frac{I(I+2mr^2)}{(I+mr^2)^2} = \hat s \left(1+\frac {mr^2}{I+mr^2}\right)$

 

[haruspex]

Energy lost to friction = loss of linear KE + Gain in rotational KE

See if you can derive that from the standard starting point: total energy before = total energy after (including that lost to friction etc.)

 

[dean barry]

Thanks for that, loads to chew on, appreciate all the help.

 

[sophiecentaur]

It is true to say that :
Over the distance described above :
Energy lost to friction = loss of linear KE + Gain in rotational KE

It probably is - but the use of "loss" and "gain" terms make it hard (for me - and possibly others) to appreciate what's going on. It may be better to come up with an equation that just uses the changes in the three components of energy with plus and minus signs? A 'loss' of 'minus' energy can get your brain in a twist and that's what happens if you try to re-arrange your equation. Start with Change in linear KE = ΔKl and let the signs do the work. -ΔKl will mean a loss of linear KE. etc. etc.
Best for you to try it yourself on a piece of paper.

 

[rcgldr]

Energy lost to friction = loss of linear KE + Gain in rotational KE

Energy lost to friction = initial linear energy - (final linear energy + final rotational energy). As mentioned previously, the initial linear energy is 1/2 m v^2, while the (final linear + angular energy) = 5/14 m v ^2, which is a loss of 1/7 m v^2. In terms of loss relative to initial energy, it's (1/7 m v^2) / (1/2 m v^2) = 2/7 of the initial energy.

 

[dean barry]

Excellent, thanks, so the model i have is :
Subtract 2/7 of the original linear KE, then calculate the linear : rotating energy split from this remainder
Calculate the final linear velocity.
Loads to go on, thanks for your time everyone.

 

[haruspex]

Excellent, thanks, so the model i have is :
Subtract 2/7 of the original linear KE, then calculate the linear : rotating energy split from this remainder
Calculate the final linear velocity.
Loads to go on, thanks for your time everyone.

That's somewhat backwards as a process.
The easiest thing to calculate from the original data is the velocity when rolling is achieved. Consider angular momentum about a point on the ground.
What is it initially? What will it be when rolling? Are there any forces which have a moment about that point?
Finding that 2/7 of the energy is lost as friction comes much later.

 

[dean barry]

Heres a thing :
I worked a start and step programme.
Start with a small (arbitrary) distance (d) traveled by the ball (say 1 Metre)
If you add 5/7 of the friction work done (m*g*k*d) to the rotation KE, then adjust the linear KE down to balance:
Then calculate the velocity of both.
(surface velocity of the ball)
You can work the distance to obtain the synchronous speed.
And 2/7 of the original KE is lost to friction as you stated.

 

[haruspex]

add 5/7 of the friction work done (m*g*k*d) to the rotation KE

On what basis?
First, the friction work done is less than m*g*k*d. Because the ball has started to rotate, the distance the two surfaces have moved in relation to each other is less than d.
Second, how do you know that 5/7 of that work went into rotational KE?
I don't understand what you are trying to do with this approach.

 

[sophiecentaur]

Heres a thing :
I worked a start and step programme.
Start with a small (arbitrary) distance (d) traveled by the ball (say 1 Metre)
If you add 5/7 of the friction work done (m*g*k*d) to the rotation KE, then adjust the linear KE down to balance:
Then calculate the velocity of both.
(surface velocity of the ball)
You can work the distance to obtain the synchronous speed.
And 2/7 of the original KE is lost to friction as you stated.

You can only say how things will end up and not what is happening during the process. With a high or low correction coefficient, the overall answer is the same. It's only when there is no more energy lost to friction that the 2/7 thing applies.
To find this distance, you just need to know the coefficient for friction and that will tell you how long / far before the slipping ends.

 

[dean barry]

Im trying to build a behaviour model for the process.
The 5/7 was a stab in the dark, i had attributed 50:50 friction force distribution but it didnt work, but the 2/7 losses thing fell into place when i did.
I had assumed full friction force throughout.
But if at the outset full friction force applies and then diminishes to zero at synchronous speed:
The work done by friction in this distance is ( friction force / 2 ) * distance.
The remaining KE (at synchronous speed) = 5/7 of the original
Ill build the model using these parameters and see how it looks.
Thanks for your patience.

 

[dean barry]

According to my model, modified as described:
(m = 10 kg, g = 9.81 m/s/s, k = 0.1, r = 0.1, initial v = 10 m/s)
distance to synchronous speed = 28.57 m

 

[sophiecentaur]

Im trying to build a behaviour model for the process.
The 5/7 was a stab in the dark, i had attributed 50:50 friction force distribution but it didnt work, but the 2/7 losses thing fell into place when i did.
I had assumed full friction force throughout.
But if at the outset full friction force applies and then diminishes to zero at synchronous speed:
The work done by friction in this distance is ( friction force / 2 ) * distance.
The remaining KE (at synchronous speed) = 5/7 of the original
Ill build the model using these parameters and see how it looks.
Thanks for your patience.

This is the wrong way round. The friction force will be highest (static friction) when the relative speeds between periphery and ground is zero. Whilst there is slippage, the friction force could well be less (sliding friction) but it's easier to assume there's no change. The fact that the force is constant allows you to assume that the angular acceleration is constant in the calculations. If the force diminished to zero, the ball would never stop slipping - just getting slower and slower at an ever slower rate.

 

[dean barry]

Thanks sophie, I've a better understanding of this problem, though a movie would be a great help.
Appreciate all the assistance.

 

[sophiecentaur]

Thanks sophie, I've a better understanding of this problem, though a movie would be a great help.
Appreciate all the assistance.

The idea is to try to run the movie in your head! So many simulations are so slick that it's easy to miss the key issues in the sensory overload. The secret is always to apply the 'rules', like Momentum Conservation. Some personal solutions can ignore those rules and lead you up 'that' creek with no paddle. I'm glad that you're sticking with it! nYou will get all the way, eventually.

 

[dean barry]

OK sophie, thanks for your help.
I confess to being somewhat overwhelmed by the response.
Im baffled by your previous post : Friction force is highest when the relative speed is zero (ie synchronous speed)
I assumed that at that point only rolling resistance was present.
Anyhow, ball enters stage left at given constant velocity (ill use 10 m/s) with rotation zero.

Its sliding so linear retardation takes place (not through translation), as it would do in the case of a block sliding.
A torque is applied also so rotational acceleration takes place.

My model assumes that the ball actually acts like part block and part ball, are you with me ?

 

[sophiecentaur]

OK sophie, thanks for your help.
I confess to being somewhat overwhelmed by the response.
Im baffled by your previous post : Friction force is highest when the relative speed is zero (ie synchronous speed)
I assumed that at that point only rolling resistance was present.
Anyhow, ball enters stage left at given constant velocity (ill use 10 m/s) with rotation zero.

Its sliding so linear retardation takes place (not through translation), as it would do in the case of a block sliding.
A torque is applied also so rotational acceleration takes place.

My model assumes that the ball actually acts like part block and part ball, are you with me ?

Yes, of course you are right that the force is zero when there's no relative movement (I guess that would be an instantaneous value) but, when there is movement, I can't see how the force can be any less than the limiting friction as the relative speed is non zero. This assumes that sliding friction is no different from static friction, of course. My reason for saying this is that I can't think what other value it could take than μmg. When a block slides along a surface, this force acts all the time. If μ is high then the acceleration is high (negative). If, as you suggest, the force gets less and less, then you could have the situation where the ball never stops slipping. Is that possible / likely/ according to experience? I am considering the most ideal case here.

 

[dean barry]

Have you any comments on the friction loss = 2/7 of the original KE ?

 

[sophiecentaur]

Actually, using conservation of angular momentum, I find that the velocity will drop to 5/7 of original.
Initial L = mrv
Final L = mrv' + 2mv'/5
etc. to give v' = 5mv/7
Which means a loss of 2/7 of velocity. That much has to be right, I think, as it's so straightforward (?).
To find loss of KE:
Original KE = mv²/2
Final KE = Linear KE + angular KE
= m( (5v/7)² + ((2X5)v/(5X7))²)/2
and I can't find anything wrong with that either.
And that suggests that the final KE is 29/49 of original
So the 2/7 loss answer really doesn't seem to fit.

 

[dean barry]

Thanks sophie, ill be back 12 noon GMT tommorow.