# Ball Rolling Without Slipping On A Turntable

• anon11
In summary, the problem is about a ball rolling on a turntable without slipping. The ball is constrained to move in the x-y plane and has an initial velocity due to the turntable's rotation. The problem involves finding the ball's motion in terms of (x,y) for a function of time. Various equations and concepts such as angular momentum, velocity, and rotation are used to solve the problem. The final answer involves equations for simple harmonic motion, but there may have been a mistake in the conversion to Cartesian coordinates.
anon11
Can someone please help me out with this tough problem?

1. Homework Statement

A ball rolls on a turntable without slipping describe the balls motion in terms of (x,y) for a function of time. (The turntable spins at a constant rate)

(This all the information that the professor gave us.)

## The Attempt at a Solution

My professor said the answer has 7/5 in it somewhere. Can someones please help? I'm really confused about this problem.Here are some sources that helped me get my answer:
http://www.warrenweckesser.net/pubs/AJP_BallRolling.pdf
http://www.tyoma.com/plain/science/papers/14/ball.pdf

I'm not quite clear on what you are given.
It looks like a turntable of unspecified radius is rotating with angular speed ##\Omega##.

Choosing rectangular coords (z axis, pointing upwards, as the axis of rotation and the origin on the upper surface of the turntable) - so ##\vec r(t)## is the position of the point of contact at time ##t##, then: ##\vec r(0) = (r,0,0)^t## (chose x-axis to be through the initial position of the ball.) The ball is constrained to move in the x-y plane in contact with the surface ... it's initial velocity wrt the z axis is not given, but it is rotating with angular velocity ##\vec \omega## which depends on it's position (because it does not slip).

How are you thinking about the physics here? (I cannot, confidently, read your handwriting.)
ie. What stops the ball from just staying where it is, can't it just sit there and spin?
If the ball moves across the turntable - isn't it's angular momentum coupled to that of the turntable?

Z (k) points out perpendicular to the turntable, X (i) points out to the right, Y (j) points out 90 degree from x (coordinate system is fixed in space). The spherically symmetric ball of mass “m”, moment of inertia “I” about any axis through its center, and radius “a”, rolls without slipping and without dissipation on a horizontal turntable (so frictional forces act on the ball at its point of contact with the turntable, but do no work). The turntable is rotating about the vertical z-axis at constant angular velocity Ωzˆ. (radius of turntable is unspecified)I apologize for my poor hand writing. I'll tell you how I attempted to solve the problem.

In part A of my work I determined that angular momentum is conserved about the moving point P. (Point P being the point on the ball in contact with the turntable.) If I'm not mistaken that would mean that the angular momentum at moving point P is coupled to that of the turntable.

In part B I broke the angular momentum into its components to examine how the ball spins:(angular momentum at Center of Mass)=(moment of inertia of a ball)*(angular velocity) or
hcm=IW

ν= (2/5)ma^2*<wxi + wyj + wzk> (angular momentum at Point P)=(angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)

hp= hcm+ a x p

= hcm + a x mv

= hcm + a x m<vxi + vyj>

= hcm + <-mavyi + mavxj>Therefore,

(2/5)ma^2*wx - mavy = C1 C=constant

(2/5)ma^2*wy + mavx = C2

Wz=C3In part C I acknowledged that because the ball doesn’t slip “The velocity of point P on the turntable” and the “velocity of point P on the ball” will be equal to each other. I used this knowledge to break up the equation and eliminate the angular velocity components so that we can solve for x and y positions

Ωk x r = v + w (cross product) (-ak) Ωk= constant rotation about the z axis

My final answer was:
x=x
o + RCos((2/7)Ωt)
y=yo + RSin((2/7)Ωt)

My professor said this was wrong and he let me take a glace at his answer. His answer had 7/5 in it somewhere.

I’m still really confused about this problem and I’m not sure if I approached it the right way. I would really appreciate it if you could give me some insight on how to solve it

Simon Bridge said:
What stops the ball from just staying where it is, can't it just sit there and spin?
In an ideal situation, that's what happens. In a real world situation, there's some drift.

In this video, the ball is seen to be moving in a circle:

Last edited:
rcgldr said:
In an ideal situation, that's what happens. In a real world situation, there's some drift.

Very interesting, now I just need to find a way to describe the balls motion in terms of (x,y). Do you have any idea how to do that?

Suppose the ball is at point (r, θ) on the disc. Its centre has tangential velocity ##r\dot\theta## and radial velocity ##\dot r##.
Since it does not slip, what does that tell you about its rotation about the tangential and radial axes?
What is its angular momentum about its centre?
What, then, is its angular momentum about the point, fixed in the ground frame, which is instantaneously the point of contact?
If you convert that to Cartesian, then set that it is a constant vector you should get equations for SHM.

I apologize for the delayed response
haruspex said:
Since it does not slip, what does that tell you about its rotation about the tangential and radial axes?
This tells me that the rotation about the tangential and radial axis will consistent with that of the spinning turntable?

haruspex said:
What is its angular momentum about its centre?
The angular momentum at the center of mass will be equal to: (moment of inertia of the ball)*(the balls angular velocity)

haruspex said:
What, then, is its angular momentum about the point, fixed in the ground frame, which is instantaneously the point of contact?
The angular momentum at fixed point on the ground would be: (angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)

haruspex said:
If you convert that to Cartesian, then set that it is a constant vector you should get equations for SHM
I attempted to do that in my previous reply. I think I may have made a mistake when converting to Cartesian Coordinates. I'm having trouble finding where I went wrong. I showed my solution to my professor and he said my answer was wrong and that it's supposed to have 7/5 in it somewhere. Can you please check my work and let my know if you see any errors.

anon11 said:
I apologize for the delayed response

This tells me that the rotation about the tangential and radial axis will consistent with that of the spinning turntable?The angular momentum at the center of mass will be equal to: (moment of inertia of the ball)*(the balls angular velocity)The angular momentum at fixed point on the ground would be: (angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)I attempted to do that in my previous reply. I think I may have made a mistake when converting to Cartesian Coordinates. I'm having trouble finding where I went wrong. I showed my solution to my professor and he said my answer was wrong and that it's supposed to have 7/5 in it somewhere. Can you please check my work and let my know if you see any errors.
I was hoping for answers to my questions in the form of equations in terms of r and θ.

haruspex said:
Since it does not slip, what does that tell you about its rotation about the tangential and radial axes?
I'm not sure how to write this in terms of r and θ, but I think rotation about the tangential and radial axis will consistent with that of the spinning turntable?

haruspex said:
What is its angular momentum about its centre?
The angular momentum at the center of mass will be equal to: (moment of inertia of the ball)*(the balls angular velocity)
Lcm = IW
Lcm = (2/5 ma2)(dθ/dt) a=radius of ball

haruspex said:
What, then, is its angular momentum about the point, fixed in the ground frame, which is instantaneously the point of contact?
The angular momentum at fixed point (point P) on the ground would be: (angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)=
Lp = Lcm + a x P
Lp = (2/5 ma2)(dθ/dt) + a x mv

haruspex said:
If you convert that to Cartesian, then set that it is a constant vector you should get equations for SHM.
After Converting I got this answer.
x=xo + RCos((2/7)Ωt)
y=yo + RSin((2/7)Ωt)
my professor said this was wrong.

anon11 said:
I'm not sure how to write this in terms of r and θ, but I think rotation about the tangential and radial axis will consistent with that of the spinning turntable?
I was maybe not clear. I meant rotation about the axes through its centre in the ##\hat r## and ##\hat \theta## directions.
We know the point of contact has velocity ##r\Omega\hat \theta##, and by definition the centre of the ball has velocity ##\vec v=\dot r\hat r+r\dot\theta\hat \theta##. So the spin of the ball about its centre, ##\vec\psi##, must satisfy ##a\vec z\times\vec\psi=r(\dot\theta-\Omega)\hat\theta+\dot r\hat r##.
There is no reason for it to have spin about a vertical axis, so ##a\vec\psi=r(\Omega-\dot\theta)\hat r+\dot r\hat\theta##. (I think I got the signs right.)
anon11 said:
The angular momentum at the center of mass will be equal to: (moment of inertia of the ball)*(the balls angular velocity)
Lcm = IW
Lcm = (2/5 ma2)(dθ/dt) a=radius of ball
We need it as a vector, so ##\vec L_{cm}=I_{cm}\vec\psi##.
anon11 said:
The angular momentum at fixed point (point P) on the ground would be: (angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)=
Lp = Lcm + a x P
Lp = (2/5 ma2)(dθ/dt) + a x mv

Vectorially, ##\vec L_p=\vec L_{cm}+ma\hat z\times\vec v=\frac 25ma(r(\Omega-\dot\theta)\hat r+\dot r\hat\theta)+ma\hat z\times(\dot r\hat r+r\dot\theta\hat \theta)##
##=\frac 25ma(r(\Omega-\dot\theta)\hat r+\dot r\hat\theta)+ma(\dot r\hat \theta-r\dot\theta\hat r)##
##=ma(\frac 15(2r\Omega-7r\dot\theta)\hat r+\frac 75\dot r\hat \theta)##
anon11 said:
After Converting I got this answer.
x=xo + RCos((2/7)Ωt)
y=yo + RSin((2/7)Ωt)
my professor said this was wrong.
Converting will not immediately give the trajectory, of course. You have to use the constancy of ##\vec L_p##.
##r(2\Omega-7\dot\theta)r\cos(\theta)-7\dot r\sin(\theta)=c_x##
##7\dot r\cos(\theta)+(2\Omega-7\dot\theta)r\sin(\theta)=c_y##
Which gives me a circle with angular frequency ##\frac 27\Omega##, same as you get.

haruspex said:
Which gives me a circle with angular frequency ##\frac 27\Omega##, same as you get.
This corresponds with the data shown near the end of the second video of my prior post:

$$\omega_{turntable} = 8.477 \ radians / second$$
$$\omega_{ballpath} = 2.297 \ radians / second$$
$$\frac{2.297}{8.477} \ \approx \ \frac{2}{7}$$

I still don't understand how my professor got ##\frac 7 5##. Do you think he could have made a mistake or is there maybe some alternative way of solving this problem?

anon11 said:
I still don't understand how my professor got ##\frac 7 5##. Do you think he could have made a mistake or is there maybe some alternative way of solving this problem?
You and I independently got 2/7, and rcgldr provides experimental confirmation.

I'm sorry, can you please explain step by step how you got your final answer (including the conversion). I'm still having trouble trying to understand all this.

anon11 said:
I'm sorry, can you please explain step by step how you got your final answer (including the conversion). I'm still having trouble trying to understand all this.
Perhaps it would be simpler if you were to explain exactly how you got 2/7.

The article from the original post third link (solution to week 21 problem, ball on turntable) shows the math involved and concludes with:

"Therefore, in view of eq. (4), we see that the ball undergoes circular motion, with a frequency equal to 2/7 times the frequency of the turntable."

haruspex said:
Perhaps it would be simpler if you were to explain exactly how you got 2/7.
I learned from the links that I have listed in my original post, and tried to follow what they did to the best of my ability. I still kind of confused with the process though. I can understand these steps.
1. Determining angular momentum at center of mass
2. Determining angular momentum at moving the point, fixed on the ground frame.
however I don't understand how you got from this:
haruspex said:
⃗Lp=⃗Lcm+ma^z×⃗v=25ma(r(Ω−˙θ)^r+˙r^θ)+ma^z×(˙r^r+r˙θ^θ)L→p=L→cm+maz^×v→=25ma(r(Ω−θ˙)r^+r˙θ^)+maz^×(r˙r^+rθ˙θ^)\vec L_p=\vec L_{cm}+ma\hat z\times\vec v=\frac 25ma(r(\Omega-\dot\theta)\hat r+\dot r\hat\theta)+ma\hat z\times(\dot r\hat r+r\dot\theta\hat \theta)
=25ma(r(Ω−˙θ)^r+˙r^θ)+ma(˙r^θ−r˙θ^r)=25ma(r(Ω−θ˙)r^+r˙θ^)+ma(r˙θ^−rθ˙r^)=\frac 25ma(r(\Omega-\dot\theta)\hat r+\dot r\hat\theta)+ma(\dot r\hat \theta-r\dot\theta\hat r)
=ma(15(2rΩ−7r˙θ)^r+75˙r^θ)=ma(15(2rΩ−7rθ˙)r^+75r˙θ^)=ma(\frac 15(2r\Omega-7r\dot\theta)\hat r+\frac 75\dot r\hat \theta)

to this:

haruspex said:
r(2Ω−7˙θ)rcos(θ)−7˙rsin(θ)=cxr(2Ω−7θ˙)rcos⁡(θ)−7r˙sin⁡(θ)=cxr(2\Omega-7\dot\theta)r\cos(\theta)-7\dot r\sin(\theta)=c_x
7˙rcos(θ)+(2Ω−7˙θ)rsin(θ)=cy7r˙cos⁡(θ)+(2Ω−7θ˙)rsin⁡(θ)=cy7\dot r\cos(\theta)+(2\Omega-7\dot\theta)r\sin(\theta)=c_y

## 1. How does a ball roll without slipping on a turntable?

When a ball is placed on a turntable, it experiences two types of motion: rotational motion due to the spinning of the turntable, and translational motion due to the force of gravity. In order for the ball to roll without slipping, the point of contact between the ball and the turntable must remain stationary, while the rest of the ball rotates around it. This is achieved through a combination of friction and the ball's own inertia.

## 2. What is the relationship between the radius of the ball and the speed of rotation of the turntable?

The speed of rotation of the turntable and the radius of the ball have a direct relationship. As the radius of the ball increases, the speed of rotation required for it to roll without slipping also increases. This is because a larger radius means the ball has to cover a greater distance in the same amount of time, requiring a higher rotational speed to achieve this.

## 3. How does the mass of the ball affect its ability to roll without slipping on a turntable?

The mass of the ball does not have a significant effect on its ability to roll without slipping on a turntable. As long as the ball has enough mass to overcome the force of friction, it will be able to roll without slipping. However, a heavier ball may require a higher rotational speed to achieve this.

## 4. What factors can cause a ball to slip on a turntable?

There are several factors that can cause a ball to slip on a turntable, including a low coefficient of friction between the ball and the turntable, a high rotational speed of the turntable, and an uneven or rough surface on either the ball or the turntable. Additionally, any external forces acting on the ball, such as wind or vibrations, can also cause it to slip.

## 5. How does the direction of rotation of the turntable affect the ball's motion?

The direction of rotation of the turntable does not affect the ball's ability to roll without slipping. As long as the point of contact between the ball and the turntable remains stationary, the ball will continue to roll in the same direction regardless of the direction of rotation of the turntable.

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