# How Long Does It Take for a Bowling Ball to Stop Sliding and Start Rolling?

• kingsmaug
In summary, the ball moves with a constant velocity down the lane until it begins to roll without slipping. Once it begins to roll without slipping, the rotational kinetic energy is less than the translational kinetic energy.
kingsmaug

## Homework Statement

A spherical bowling ball with mass m = 4.3 kg and radius R = 0.102 m is thrown down the lane with an initial speed of v = 8.2 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.29. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.

magnitude of angular acceleration during sliding: 69.66 rad/s^2
magnitude of linear acceleration during sliding: 2.84 m/s^2
how long to roll without slipping: 0.824 s
length of slide before rolling: 5.79 m
and after it begins to roll without slipping, the rotational kinetic energy is less than the translational kinetic energy.

What is the magnitude of the final velocity?

KE=1/2mv^2
Erot=1/2Iw^2
w=v/r

## The Attempt at a Solution

I assumed that the KE of the ball immediately before hitting would be equal to the sum of the final KE and Erot.

KE1=KE2+Erot

1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2)

After mathing, I solved that the final velocity is 1.30559 m/s, but that isn't right.

You can apply the conservation of total energy if you can properly account for the changes (transfers) of energy during the process from start to finish.

*facepalm* Energy loss due to friction.

So,

KEi = KEf + Ertof - Efric

1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2) - uk(N)(d)

Which when solved gives me vf=1.59 m/s, which is also wrong.

kingsmaug said:
KEi = KEf + Ertof - Efric

1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2) - uk(N)(d)

Is your rotational kinetic energy correct? (do the units makes sense?)

Units, yes. But I just realized that I left out the r^2 term from I.

So the rotational energy is 1/2*I*w^2, and w=v/r

That turns to 1/2*I*(v/r)^2

But I suppose it doesn't entirely matter anymore. I've run out of attempts on the problem.

kingsmaug said:
But I suppose it doesn't entirely matter anymore. I've run out of attempts on the problem.

Possibly,... if you never have to solve a problem like that in the future.

The friction decelerates the linear velocity according to v=Vo-at. If you know the value of a, and also the time of skidding, t, why do you not use the formula v=Vo -at to get the final linear velocity?

Last edited:
The easiest way to solve this is by considering angular momentum. The trick is to choose a reference axis such that you don't care about the frictional force.

The magnitude of the final velocity can be solved for by using v=vi-ugt, where t=(2vi)/(7ug).
Substitute t into the formula and solve for v.
Remember, vi is given to you in the problem.

legend:
u=mu
g=gravity
t=time
vi=initial velocity
v=what you're solving for

physics09 said:
The magnitude of the final velocity can be solved for by using v=vi-ugt, where t=(2vi)/(7ug).
Substitute t into the formula and solve for v.
Remember, vi is given to you in the problem.

legend:
u=mu
g=gravity
t=time
vi=initial velocity
v=what you're solving for
As I posted, there's no need to find the time if all you care about is final velocity. Just use conservation of angular momentum.

if you ignore frictional losses, then if you assume the ball is a solid homogenous sphere, the final ratio of linear KE to rotational KE will be:
5/7 : 2/7
This totaled will match the original linear KE

if you include friction converted to heat:
As linear deceleration and rotational acceleration are constant, then the driving forces are also deemed to be constant, if you work these out from the figures, the rest of the data should fall into place.
Use time elapsed as a guide, calculate the point in time where diminishing linear velocity and increasing rotational surface velocity are equal.
(synchronous speed)

dean barry said:
if you ignore frictional losses,
Why would you do that? It will clearly produce a wrong answer. In particular, it will violate other conservation laws.

im going from the given numbers
the time given (0.824 s) and the derived distance (5.79 m) are correct if you use the method i outlined (post 12)
I did it by trial and error based on elapsed time and you can then figure the final velocity using:
elapsed time (0.824 s)
initial velocity (8.2 m/s)
and linear deceleration rate ( - 2.84 m/s^2 )
by using the appropriate Newtons equation of motion
with the final velocity you can calculate the linear KE and then use the 5/7 : 2/7 rule to find the rotational KE

dean barry said:
im going from the given numbers
the time given (0.824 s) and the derived distance (5.79 m) are correct if you use the method i outlined (post 12)
I did it by trial and error based on elapsed time and you can then figure the final velocity using:
elapsed time (0.824 s)
initial velocity (8.2 m/s)
and linear deceleration rate ( - 2.84 m/s^2 )
by using the appropriate Newtons equation of motion
with the final velocity you can calculate the linear KE and then use the 5/7 : 2/7 rule to find the rotational KE
Or just use conservation of angular momentum about a point on the ground:
##m v_i R = m v_f R + I v_f / R = m v_f R(1+\frac 25)##
Note that the mass, radius and coefficient of friction are all irrelevant.

## 1. What is a slipping bowling ball?

A slipping bowling ball is a term used to describe a bowling ball that is not rotating properly on the fingers of the bowler. This can be caused by a variety of factors, such as incorrect finger placement or inadequate grip on the ball.

## 2. What are the potential consequences of a slipping bowling ball?

A slipping bowling ball can lead to a decrease in accuracy and power, resulting in a lower score. It can also cause strain on the bowler's fingers and wrist, leading to potential injuries.

## 3. How can a bowler prevent a slipping bowling ball?

To prevent a slipping bowling ball, a bowler should ensure that their fingers are placed correctly in the finger holes, and that they have a firm and consistent grip on the ball. Using a wrist support or a rosin bag can also help improve grip.

## 4. Can the weight of a bowling ball affect its tendency to slip?

Yes, the weight of a bowling ball can play a role in its tendency to slip. A heavier ball may be more difficult for a bowler to control, while a lighter ball may not provide enough momentum for a strong release.

## 5. What should a bowler do if their bowling ball continues to slip?

If a bowler continues to experience a slipping bowling ball, they may need to adjust their grip or finger placement, or try using a different ball with better finger holes. It may also be helpful to consult with a coach or professional to identify and correct any underlying issues.

• Introductory Physics Homework Help
Replies
8
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
60
Views
829
• Introductory Physics Homework Help
Replies
24
Views
276
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
4K
• Introductory Physics Homework Help
Replies
32
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K