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

  1. Mar 17, 2005 #1
    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:

    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?
  2. jcsd
  3. Mar 17, 2005 #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.
  4. Mar 17, 2005 #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)?
  5. Mar 17, 2005 #4


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    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.

    reilly Atkinson
  6. Mar 17, 2005 #5
    Is there ever a discontinuity in the ball's motion?
  7. Mar 17, 2005 #6
    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.
  8. Mar 18, 2005 #7

    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

    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.
  9. Mar 18, 2005 #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.
  10. Mar 19, 2005 #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?
  11. Mar 19, 2005 #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.
  12. Mar 19, 2005 #11
    Has anyone considered introducing the simplification of zero friction?
  13. Mar 19, 2005 #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.
  14. Mar 20, 2005 #13
    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!
  15. Mar 20, 2005 #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).
  16. Mar 20, 2005 #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.
  17. Mar 20, 2005 #16
    Why didn't I think of it?
  18. Mar 20, 2005 #17
    Crosson: Is that what differential equations is pretty much all about?
  19. Mar 21, 2005 #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.
  20. Mar 22, 2005 #19

    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.
  21. Mar 22, 2005 #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 dont 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:

    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:
    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)

    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.

    b=-1/k, because the normal is perpendicular to the resultant.

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

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

    now we insert these formulas into the force formula for y:
    this y is not same as above though, so i will call it f:
    inserting 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: Mar 22, 2005
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