Mechanics problem: a ball rolling off the edge of a table

In summary: The ball is constantly accelerating in the y direction, but the x direction is a little bit more complex.In summary, the conversation discusses setting up equations for the movement of a ball that is placed on the edge of a table and then rolls down due to gravity. The expert suggests using Newton's 2nd law and solving differential equations to find the force acting on the ball at any given coordinate. They also mention the constraints and the use of calculus in solving this problem. The conversation ends with discussing the acceleration of the ball in both the x and y directions.
  • #1
ribod
14
0
I'm requesting help for a physics problem I have made up.

Let's say we put a ball on the edge of a table, with the center of the ball being a teeny weeny bit outside the edge of the table. In other words, if we have a coordinate system, the edge of the table is at 0,0, the center of the ball is an extremely small bit to the right of 0,r (where r is the radius of the ball).

It looks something like this:
''''''''''''''''''''''''ball
_________(_'''')
table'''''edge^

What will happen now is that the ball will roll down over the edge, to the right, because of gravity.

My question is, how can I set up two functions, for the movement of the center of the ball -- x(t) and y(t), where t is the time -- only when the ball is rolling over the edge.

My thinking is that the movement of the ball will follow the shape of the ball. At any given coordinate, the force affecting the ball can be calculated simply by taking the gravity vector downwards, the normal vector pointing from where the ball touches the edge to the center of the ball, and from these get the resultant vector. This resultant will be a tangent to the ball's circular shape, where the ball touches the edge. This means that the acceleration will change over time, and thus is not constant.

Let's say gravity acceleration is constant G, and radius of the ball is r, the center of the ball is x,y, and the coordinates where the ball touches the edge of the table is h,H. We ignore air resistance, and such things that can be ignored.

I can not use for example s=vt, because v is not constant. I cannot use s=ut+at^2/2, because acceleration is not constant either. so...

How will the ball move with time? How do I do the equations?
 
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  • #2
The answer comes from looking more deeply at Newton's 2nd law:

F = ma

Write this as:

[tex] \Sigma F_x = \frac{d^2 X(t)}{dt^2} [/tex]

[tex] \Sigma F_y = \frac{d^2 Y(t)}{dt^2} [/tex]

Now you need to find the force as a function of position and velocity!

[tex] \Sigma \vec{F(\vec{r},\vec{v})} [/tex]

Using Newton's 2nd, you will have a set of (coupled) differential equations.

[tex] F_x (X,Y,V_x,V_y) = \frac{d^2 X(t)}{dt^2} [/tex]

[tex] F_y (X,Y,V_x,V_y) = \frac{d^2 Y(t)}{dt^2} [/tex]

Then you have to know how to solve differential equations.
 
  • #3
Ok could you explain it more detailed please?

First of all what is d? Then how do you get ma=d^2X(t)/(dt^2)?
 
  • #4
The keys are the constraints. As you have set it up, there is gravity acting on the CM, and a force exerted by the table edge on the ball. (Often this problem is done starting with the ball rolling on the table toward the edge.) Ask yourself, does the ball roll down the edge as it moves under gravity? This is, more or less, a standard constraint problem, best done with a Lagrangian approach and Lagrange multipliers.

Regards,
reilly Atkinson
 
  • #5
Is there ever a discontinuity in the ball's motion?
 
  • #6
First of all what is d? Then how do you get ma=d^2X(t)/(dt^2)?

Think of "d" like you think a "change in", as in velocity = dx/dt. dx/dt is called the "derivative" of x with respect to t, and it means "the instantaneous rate of change in x".

I am not trying to sound mean, but you cannot solve this problem, because you have not studied calculus.
 
  • #7
ribod said:
I'm requesting help for a physics problem I have made up.

Let's say we put a ball on the edge of a table, with the center of the ball being a teeny weeny bit outside the edge of the table. In other words, if we have a coordinate system, the edge of the table is at 0,0, the center of the ball is an extremely small bit to the right of 0,r (where r is the radius of the ball).

It looks something like this:
''''''''''''''''''''''''ball
_________(_'''')
table'''''edge^

What will happen now is that the ball will roll down over the edge, to the right, because of gravity.

My question is, how can I set up two functions, for the movement of the center of the ball -- x(t) and y(t), where t is the time -- only when the ball is rolling over the edge.

My thinking is that the movement of the ball will follow the shape of the ball. At any given coordinate, the force affecting the ball can be calculated simply by taking the gravity vector downwards, the normal vector pointing from where the ball touches the edge to the center of the ball, and from these get the resultant vector. This resultant will be a tangent to the ball's circular shape, where the ball touches the edge. This means that the acceleration will change over time, and thus is not constant.

Let's say gravity acceleration is constant G, and radius of the ball is r, the center of the ball is x,y, and the coordinates where the ball touches the edge of the table is h,H. We ignore air resistance, and such things that can be ignored.

I can not use for example s=vt, because v is not constant. I cannot use s=ut+at^2/2, because acceleration is not constant either. so...

How will the ball move with time? How do I do the equations?


Force the ball to not slip on the corner as it rolls off.

The equations for the center of the balss become
x = cos(theta), y=sin(theta) where the origin is on the corner of the
table.

As for how it moves with time, you need the above posts to help you out
with the forces. It's not a simple problem.
 
  • #8
Any x component the ball will experience will be a result of the normal force from the table pushing the ball out of the table's way as the ball falls. This will be the most diffulct part to figure out, though it will still be a function of time, since the ball drops at a constant acceleration.
x(t) = some function of y(t) where
y(t) = g*sin(theta)

Notice sin(theta) will not be constant since the ball experiences different forces as it passes the table.

For every instant t where the ball is pulled downthe table exerts a force equal in magnitude to the force imparted on it by gravity. There is a torque applied to the ball, it will spin around its CM as it falls off the table. As to how to calculate each of these components, I'm not really sure how you would go about doing so.
 
  • #9
Crosson I know what a derivate is. If it would say dx/dt i would have known it was the derivate. However I don't remember any use of d squared. It was some years ago I studied this.

I found your formula online however, and I want to make sure if there is no algebraical solution to my problem. I have to use numerical method?
 
  • #10
What is it about the motion that is important?

The method I described will give you all the information about the balls position at any time, but if all you want to know is the speed it falls with, or how many times it spins, that problem is easier.
 
  • #11
Has anyone considered introducing the simplification of zero friction?
 
  • #12
crosson as I understand it you showed that s''=v'=a ? this doesn't make a function of time.

Can anyone solve the problem? It might be trickier than you think if you don't really solve it.
 
  • #13
s''=v'=a ? this doesn't make a function of time.

That's why I said more than that. Realizing that acceleration is defined as a derivative does not solve problems, but Newton's law does:

F = ma

[tex]F(x,v) = m \frac{d^2 X}{dt^2}[/tex]

Can you find the forces as a function of position and velocity?

Let me give you an example, suppose we are talking about a spring mass setup. Then the force due to the spring is proportional to how stretched out it is:

F = - k*x(t)

So to determing the motion we just use:

[tex]F(x,v) = m \frac{d^2 X}{dt^2}[/tex]

[tex]-kx = m \frac{d^2 X}{dt^2} [/tex]

This is a functional equation (where the function x(t) is the unknown). We can't solve for x(t) using algebra, because we have to deal with the second derivative of x. We need a function x, -k times which will equal m times its second derivative. It turns out the solution is:

x(t) = A cos(wt) + B sin(wt)

Springs are oscillators!
 
  • #14
yes i can make a function of position, F(y), or F(x), but i have no idea how you get from -kx=m*a to x(t)=A cos(wt) + B sin (wt) (let alone where did A and B come from).
 
  • #15
Good point, why do something at all if I am not going to do it right?


[tex]-kX(t) = m \frac{d^2 X(t)}{dt^2} [/tex]

look at what this equation says: We need a function x, which is the negative of its second derivative.

Most functions do not have this property. For example, 5t^3 is not equal to the negative of its second derivative. Neither is e^t, ln(t), t^(-2/3). But sin(t) and cos(t) are the negatives of their second derivatives!

So they are the "solution" to the above differential equation. (We have uniqueness theorems that tell us these are the only solution.)

I think this problem is one of the most beautiful of them all, judged by its power and simplicity.
 
  • #16
Crosson
I think this problem is one of the most beautiful of them all, judged by its power and simplicity.
Why didn't I think of it?
 
  • #17
Crosson: Is that what differential equations is pretty much all about?
 
  • #18
Yes, this is a differential equation. Notice we guessed the solution; applying (difficult) special techniques to solve them is what differential equations is all about.
 
  • #19
ribod said:
yes i can make a function of position, F(y), or F(x)...


What did you get? I don't know how to set up force equations at the edge of the table.

If you imagine that the ball is rolling off a table with a rounded "edge" (like a quarter-circle), and require that the ball is rolling without slipping, you can find the point at which the ball loses contact with the table in terms of the initial velocity. This is when the force of gravity exactly balances the centrifugal force, which makes the normal force zero. However, if the initial velocity is too large, the ball may lose contact immediately upon reaching the rounded part. Here is the condition that must be required for the ball to stay in contact with the rounded part for some finite time:

[tex]\frac{V_0^2}{gR} \leq 1 [/tex]

Where V_0 is the initial speed of the ball and R is the radius of the rounded edge. As R approaches zero, as in your case, no finite velocity will allow the ball to stay in contact with the edge.
 
  • #20
my reasoning, which can be illustrated easily visually by using a cd disc rolling of the table:
the corner is like a point, and does not affect the geometry of the movement. since the cd is round however, the movement will be the same as if a point is rolling off a sphere.
to apply the force is easy. at any position of the ball, or cd, with force down, the ball can't move down, so it has to move as much down as it can. and that is always the tangent line of the point where the ball touches the table. the length of this vector is calculated the same way as when a ball is rolling on a slope.

let's say g is gravity down, with the center of the ball in (0,0) in a coordinate system. that means a gravity vector ends in (0,g). geometry shows that the normal force will always point from the center of the wheel towards the point where the ball hits the edge. this gives you two coordinates to calculate the line of the normal vector. if you don't trust this you can just find out the line of the normal knowing it passes through the y-axis at x=0,y=g, and using the coordinates where the ball touches the edge. this gives us a function for the normal in the form of:
y=kx+g

that gives you two lines, gravity down and the normal, intersecting the at (0,g). now draw the resultant force line from 0.0, to where it intersects with the normal force. this coordinate will be the vector of the resulant force, at any coordinate.

to find out the value, we simply use
y=kx+g and y=bx
where b is the coefficient for the line that represents the direction of the resultant force. both y's and x's are equal of course, because we want to know where these two lines meet.

solving this:
kx+g=bx
x=g/(b-k)
gives us the force on the x axis. for y-axis you simply do:
x=(y-g)/k; x=y/b
y/b=(y-g)k; yk=yb-gb; gb=yb-yk=y(b-k)
y=gb/(b-k)

now i can simply find out how a force relates to the distance of x or y, because b and k use x and y.

the normal force's line:
k=(y-H)/(x-h) where (x,y) are the center of the wheel, and (h,H) are the coordinates of the hit point.
also:
y=kx+g;
k=(y-g)/x

b=-1/k, because the normal is perpendicular to the resultant.
b=x/(g-y)

since we know how x relates to y in this point, we make these formulas depend on only y:
y=bx; x=y/b
k=(y-g)/x=(y-g)/(y/b)=-1/b
b=-sqrt(-y)/sqrt(y-g) OR
b=sqrt(-y)/sqrt(y-g)

so:
k=-1/b= + OR - sqrt(y-g)/sqrt(-y)

now we insert these formulas into the force formula for y:
y=gb/(b-k)
this y is not same as above though, so i will call it f:
f=gb/(b-k)
inserting y:
f=g(sqrt(-y)/sqrt(y-g))/((sqrt(-y)/sqrt(y-g))-(-sqrt(y-g)/sqrt(-y)))
OR
f=g(-sqrt(-y)/sqrt(y-g))/((-sqrt(-y)/sqrt(y-g))-(sqrt(y-g)/sqrt(-y)))

...
i can edit this post later to explain some details, too tired now.
btw crosson i get it now, thanks for your help. i did find an alternate solution however, by using s=v0*at^2/2, and finding out the averege acceleration with the formula of f(y).
 
Last edited:
  • #21
I see. You mean the ball is sliding off the edge, initially at rest. I was going by the title of the post, but I should have re-read the original post.

Still, don't you think the ball will pick up a horizontal component of velocity from the edge, pushing it away from the table? Your equations don't indicate that the ball will ever lose contact with the table.
 
  • #22
this is the movement on the edge, not meaning it will never slide off. i only wanted this equation, since the movement after that is much easier to calculate.
 
  • #23
Ball On the Edge

You may want to check out:
Beeken, "Ball on the Edge", The Physics Teacher, September 2004, Volume 42, Issue 6, pp. 366-368

It includes experimental measurements as well as the modelling equations.
 
  • #24
1. The Friction is non zero otherwise the ball will not roll and simply slid down without any angular velocity about an axis passing through center of mass as other two forces passing through the C.M.

2. The friction and normal reaction both are varying.

3. If the friction is sufficiently large, not allowing the ball to slide even then the ball will leave the edge at some particular angle.

4. As friction and normal reaction both are varying it is easy to consider rotation about the edge, which is the axis of rotation.

5. At a particular angle turned (theta) with the vertical the torque about the edge is mgr sin(theta) and we can calculate the angler acceleration of the ball and hence acceleration of the center of mass of the ball which can be resolved to give component of acceleration in X and Y directions,
----------------
6. Using the law of conservation of energy we can find angular velocity of the ball about the edge as a function of theta.

7. Angular velocity can get velocity of the C.M. can be resolved in X and Y direction.

I think that much is sufficient to set equations for x and y which are rsin(theta ) and rcos(theta)
 
  • #25
In the first diagram I have attached, the black = table, the yellow = ball, the green lines are the vertical line and the line connected from the edge of table to the centre of the ball, the blue line is the tangential line of the contact point of the ball, and the red line is the resultant force acting on the falling ball.

Let θ = the angle between the 2 green lines.
There are 2 cases I will discuss here. The first is when the table just exerts normal force on the ball but no friction. The second is the case when there is a friction force in addition to the normal force.

Case 1: No friction.
It is important for us to know the direction of the forces.
In my opinion, the normal force will be always pointing in the direction along the line going to the centre (the green line)! My explanation for this is, the table will never “feel”
The component mg sinθ but only “feel” the component mg cosθ.
Hence, the component mg cosθ will be canceled out, leaving mg sinθ as the net /resultant force.
This is similar to the case in which a particle is falling down a hemisphere with initial speed = 0 at the top.
I also find that the ball will NEVER rotate; this is because there is no force acting along the blue line to give a torque about the axis passing through the centre of ball (and perpendicular to the page!).
However, the ball will not lose contact with the table initially, but at a special angle x.
We can visualize the process as in the second diagram I have attached.




Although the ball is not rolling, it still can get contact with the table! At time t, the ball “tends” to slide away the edge of table. In an extremely short duration ∆, the ball moves in the direction of mg sinθ and MEAWHILE fall down a little bit due to gravity (in this extremely small duration, it has lost the contact with the table, hence the net force = mg!)
After it fall a little bit (a very very small distance!), it will come in contact with the table again! But this time, the point in contact has changed to another point as illustrated in the diagram above!
So, what happen is: when the ball is in contact with the table, the CM of the ball will move and being accelerated in a curve such that the distance between it and the edge of the table is constant (the radius of the ball). In fact, the CM’s motion is the same as the motion of a particle falling down a hemisphere with initial speed = 0!
And, same as the particle, the ball will lose contact with the table at a special angle x, and you can get the value of x in the same way as you did for the particle!
(Hints: hen the normal force <0, i.e. when the mg <centripetal force required.!)

Case 2:
When there is friction (maybe because the edge of table has slightly sunk into the ball surface.) the friction will act to:
1) Decrease the net force; hence decrease the acceleration of CM.
2) Give the ball a torque, hence make it rotate!
3) When the friction force is large enough and constant, the contact point of the ball will
never move! This is same as when the boll is PIVOTTED at that point! In this case, the CM of the ball will rotate 90 degree around the edge of table. Its motion (acceleration and velocity) can be determined by using torque = mgRsinθ.


Hope my opinionsand ideas can help you.
Thanks,
Twukwuw.
 
  • #26
the pictures are attached.
 

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1. How does the height of the table affect the ball's velocity?

The height of the table does not directly affect the ball's velocity. However, as the ball falls off the edge of the table, it will experience a change in velocity due to the force of gravity acting on it.

2. What factors determine the distance the ball will travel before hitting the ground?

The distance the ball will travel before hitting the ground is primarily determined by the initial velocity of the ball and the height of the table. Other factors such as air resistance and the shape of the ball may also play a role.

3. How does the shape of the ball affect its motion?

The shape of the ball can affect its motion in several ways. A spherical ball, for example, will experience the same force of gravity on all sides, resulting in a more predictable and consistent motion. A non-spherical ball may have uneven weight distribution, causing it to roll or spin in a certain direction as it falls off the table.

4. How does air resistance impact the ball's motion?

Air resistance can slow down the ball's motion as it falls off the table. This is because the air molecules push against the ball in the opposite direction of its motion, creating a drag force. The amount of air resistance depends on the density of the air and the shape of the ball.

5. Can the ball's motion be accurately predicted using physics equations?

Yes, the ball's motion can be accurately predicted using physics equations such as the equations of motion and the law of conservation of energy. However, in real-world situations, there may be other factors such as air resistance and friction that can affect the ball's motion and make it more challenging to predict accurately.

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