# Bowling ball problem

• dean barry

#### dean barry

A bowling ball is released with no rotation (linear KE only) at a given speed, the friction between the ball and the lane creates a torque which tends to rotationaly accelerate the ball, robbing it of linear KE and reducing the linear speed until synchronous speed is reached.
My question is :
The rotational torque i got from : m * g * µ * r
m = mass
g = local gravity rate
µ = friction co-efficient
Is all the friction force absorbed in rotating the ball, or does part of it decelerate the ball directly ?
Thanks
Dean

Try applying conservation of energy to this problem. Before the ball touches the floor it has a certain amount of kinetic energy from its forward motion and a certain amount of kinetic energy (zero, because it hasn't started rotating yet) from its rotation. Once it is rolling across the floor in steady state it has the same total amount of kinetic energy, but now the rotational component is non-zero. What does this mean for its forward motion?

• jackwhirl
Its diminished, I am aware of the translation from linear to rotational, my problem is that : does all of the the force from sliding friction rotate the ball ?
Ive attached the word sheet i put together for reference.
Dean Barry

#### Attachments

• bowling ball.xlsx
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Force from the sliding friction acts both as a torque that spins the ball and as a linear force which decelerates the ball. The full force acts in both manners. None of it is "used up" for one purpose and thereby made unavailable for the other.

ive sent the excel sheet, sorry, here's the word doc.

#### Attachments

• bowling ball.docx
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So, the friction force = m * g * µ = 10.791 N , i attributed all of that to the torque rotating the ball, what do i do now ? half to rotation and half to linear deceleration ?
Thanks
Dean

All to rotation and all to deceleration.

Dean,
In real life some minuscule amount of the kinetic energy of a bowling ball is converted into heat energy through friction. Similarly energy is expended on pushing all those troublesome air molecules out of the way like trillions of little bowling pins. In the grand scheme of things, though, these are not worth considering.
So, the friction force = m * g * µ = 10.791 N , i attributed all of that to the torque rotating the ball, what do i do now ? half to rotation and half to linear deceleration ?
Since we are ignoring the lesser effects:
All of the energy given to the rotation is taken from the linear motion.
All of the energy taken from the linear motion is given to rotation.

As jbriggs444 stated, the force you calculated is applied equally and oppositely to each.

Once it is rolling across the floor in steady state it has the same total amount of kinetic energy, but now the rotational component is non-zero.
How can you say the kinetic energy is the same?

I can understand for, say, a ball rolling down an inclined plane. The point of contact (where the force of friction is acting) is stationary, and thus no work is being done by friction (right?).

But in this case, the point of contact is sliding along the ground until it reaches a steady state, so wouldn't you have to account for energy lost to friction?

But in this case, the point of contact is sliding along the ground until it reaches a steady state, so wouldn't you have to account for energy lost to friction?

You do if you want to calculate the exact final speed, but the correction is small. You don't if you just want to answer OP's question.

Energy conservation is the wrong way to solve this problem. Kinetic energy is not conserved. Not even approximately.

An approach based on angular momentum and linear momentum is appropriate. First, pick a reference axis about which to compute angular momentum. Two choices suggest themselves. Either choice should allow you to determine a ratio k between the rate at which linear velocity is reduced and the rate at which tangential velocity is increased.

That gives you one equation: linear-velocity-reduction = k x tangential-velocity-increase

When the final stable state has been attained, the ball's linear velocity will match its tangential velocity. That gives you a second equation:

starting-linear-velocity - linear-velocity-reduction = starting-tangential-velocity + tangential-velocity-increase.

You know the starting linear velocity and the starting tangential velocity. So that's two equations and two unknowns. Solve them.

Right gents, I am sticking with the conservation of KE applies for now.
Calculate the friction force (f).
Then :
1) calculate linear deceleration using a = f / m
2) calculate deceleration due to energy transfer

Dean Barry

The kinetic energy is not conserved as there is sliding friction.
Treat translation and rotation of the ball separately.
Your first steps are right: The force of friction is f=μmg.The translation of the ball decelerates : a = -f/m . How does the velocity change with time?
The force of friction acts with a torque τ=rf around the centre of the ball and accelerates rotation. The angular acceleration is α=τ/I (I is the moment of inertia)
How does the angular velocity change with time?
When pure rolling, the angular velocity and the translational velocity of the CM are related, how? From that you get the time when the ball starts to roll. Knowing the time, you get the velocity and angular velocity.

Right gents, I am sticking with the conservation of KE applies for now.
I think this is basically a flawed idea. If any of the energy is transferred from linear to rotational by friction then there needs to be Work Done, turning the ball and slowing its linear velocity. That work will, I think, also involve work against the friction force. I would go so far as to say that the energy dissipated through the friction will be the same as the energy transferred to rotational KE.
There is an electrical analogue to this when an uncharged Capacitor is connected across a charged one. (an old chestnut that students are often presented with) The total electrostatic energy is always less than the original, irrespective of the actual value of the resistance (of which there must always be some). For two equal value capacitors, the energy dissipated is actually half. So I would say that the actual coefficient of friction would not affect the bowling ball result. Once slipping has stopped, the tangential speed is the same as the linear speed. The simplified, ideal model with no slippage will not actually work as there will be an unacceptable infinite force at first contact.

• jbriggs444
im thinking maybe the friction force (and therefore torque) is maximum at the beginning and ramps down to zero at synchronous speed, deceleration not being at a constant rate.
im going to sit with this for now, thank you all for your help
dean

im thinking maybe the friction force (and therefore torque) is maximum at the beginning and ramps down to zero at synchronous speed, deceleration not being at a constant rate.
im going to sit with this for now, thank you all for your help
dean
When dynamic friction is the same as static friction, this is not the case and the force would be constant, giving linear acceleration. But your point is a good indication that the Energy is not conserved (a force times a distance) so that makes doing it your way much harder. As with most interaction problems, it is much easier to stick with conservation of Momentum for your calculations. It simplifies things so much that it almost seems like cheating!.

Thank you, i will ponder this.
Dean

return jan 5th

Taking the MI of a solid sphere as 2Mr2/5, it looks like the final linear velocity of the ball is 0.7 of the original. and about 30% of the original energy is lost to friction. I won't show the workings until someone comments / agrees with me. Wotcha think?

Taking the MI of a solid sphere as 2Mr2/5, it looks like the final linear velocity of the ball is 0.7 of the original. and about 30% of the original energy is lost to friction. I won't show the workings until someone comments / agrees with me. Wotcha think?
I agree. I find the final velocity and the final energy to be 5/7 of their respective initial values (using the MI of a solid sphere).

I would go so far as to say that the energy dissipated through the friction will be the same as the energy transferred to rotational KE.
Interesting idea! I would have never noticed this without doing the math.

energy dissipated through the friction will be the same as the energy transferred to rotational KE.
As ehild pointed out, the total energy is reduced via conversion to heat because of the sliding friction. For a uniform solid sphere, the loss to heat is 1/7 m v_initial^2 (where v_initial is the initial sliding without rotating velocity) . The math is shown in this old thread:

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thanks all, I am going to sit with equal losses to friction and translation, thanks for your time.
dean barry

thanks all, I am going to sit with equal losses to friction and translation, thanks for your time.
dean barry
Not sure what you mean by that.
Of the original KE, 2/7 will be lost to friction, 10/49 will end up as rotational energy, and the remaining 25/49 will be linear KE.

Of the original KE, 2/7 will be lost to friction ...
From the previous thread on this, the calculated result was final energy = (1/2) (5/7)^2 linear energy + (1/2) (2/5) (5/7)^2 rotational energy = 25/98 linear energy + 10/98 rotational energy = 35/98 = 5/14, so 2/14 == 1/7 was lost to friction.

update - that should be 1/7 m v^2 lost to friction. As haruspex pointed out, since the original energy was 1/2 m v^2, then the fraction of initial energy lost to friction is 2/7 = ((1/7 m v^2) / (1/2 m v^2) .

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2/14 == 1/7 was lost to friction.
No, that's 2/14 of mv2. The original energy was mv2/2, of which 2/7 is lost to friction.

No, that's 2/14 of mv2. The original energy was mv2/2, of which 2/7 is lost to friction.
Thanks I updated my previous post. It wasn't clear to me that this was a reference to the fraction of the original energy, as opposed to the actual loss of kinetic energy (for example 1/7 joules lost to friction when the initial energy was 1/2 joule).

The difference in speed between the ground and the ball means a constant sliding friction is in force ?
This then suddenly plummets to zero at synchronous speed ?

after the ball reaches synchronous speed , the friction force still continues to decelerate the ball and the ball decelerates until it reaches 0 velocity

At synchronous speed, constant linear speed is achieved (momentarily), and then rolling resistance then decelerates the ball to zero ?

At synchronous speed, constant linear speed is achieved (momentarily), and then rolling resistance then decelerates the ball to zero ?
but there is a catch here ... the bowling ball is always decelerating , it's not like the lost KE = gained rotational KE ... if you assume it then you don't have any force working when you have gained the synchronous speed , part of the total KE is always transforming into heat energy and also note that the ball doesn't stop rotating before its linear velocity becomes 0 .. so after the synchronous speed is gained , friction force works against both types of motion

Thanks, ill ponder that, back tommorow AM
Dean

The difference in speed between the ground and the ball means a constant sliding friction is in force ?
This then suddenly plummets to zero at synchronous speed ?
More or less. Similarly with a box sliding along a floor. At the moment it comes to rest, friction falls to zero, in principle. In reality, no body is completely rigid, so the drop to zero does take a short time.
after the ball reaches synchronous speed , the friction force still continues to decelerate the ball and the ball decelerates until it reaches 0 velocity
The friction only slows the linear motion of the ball (and accelerates the rotational motion) until rolling contact is achieved. At this point frictional force has become zero. In the absence of rolling resistance it would then just keep rolling steadily.
At synchronous speed, constant linear speed is achieved (momentarily), and then rolling resistance then decelerates the ball to zero ?
If there's rolling resistance there's no reason why the linear deceleration should be momentarily zero, though it will drop substantially compared with sliding.

OK, thanks all for your time on this, I am going to summarise as follows :
Ball enters sliding at given speed,
Friction force (m*g*µ) is split evenly between direct linear deceleration due to sliding friction ( a = f / m ), and torque increasing rotation (which also produces linear deceleration).
I now need the time and distance to synchronous speed.

Ball enters sliding at given speed,
Friction force (m*g*µ) is split evenly between direct linear deceleration due to sliding friction ( a = f / m ), and torque increasing rotation (which also produces linear deceleration).
No, you've been told that's wrong. It is not split; it applies fully to both simultaneously.

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 ?