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Old Mar 8th 2011, 06:33 PM   #1
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Ball rolling under gravity

A ball rolls under gravity inside a smooth bowl with circular cross section and inner radius a. Show that the ball can roll in a horizontal circle at a height h above the bottom of the bowl provided:

v^2=g*h(2*a-h/a-h)

OK, well I'm not certain at all how to begin this problem. The only one of the values I know is g=10 m/s^2.

I suspect this has something to do with a conical pendulum?

I know there is no vertical motion so i can ignore that for a start. The only force I can reason to acting on the ball is gravity

The ball has to be moving at some angle theta to the vertical and that at the given height h the radius is a

Any suggestions where to start?
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Old Mar 8th 2011, 07:29 PM   #2
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Originally Posted by GrahamJamesK View Post
A ball rolls under gravity inside a smooth bowl with circular cross section and inner radius a. Show that the ball can roll in a horizontal circle at a height h above the bottom of the bowl provided:

v^2=g*h(2*a-h/a-h)

OK, well I'm not certain at all how to begin this problem. The only one of the values I know is g=10 m/s^2.

I suspect this has something to do with a conical pendulum?

I know there is no vertical motion so i can ignore that for a start. The only force I can reason to acting on the ball is gravity

The ball has to be moving at some angle theta to the vertical and that at the given height h the radius is a

Any suggestions where to start?
I would first like to make a comment. It is going to be next to impossible to lead you through without essentially "giving away" the solution. I urge you to carefully study this and make sure you can derive every equation below. Else you will gain nothing.

This is going to be a pain in the neck without a decent diagram, but I'm afraid I'm not up to sketching it. (And my scanner is down.)

So. Let's assume our spherical bowl has its minimum at the origin (with the usual +x and +y directions). I am going to arbitrarily pick the problem to be in the xy plane...any cross-section of the problem will do. Our segment of the sphere in the xy plane is given by x^2 + (y - a)^2 = a^2. I am choosing to work in the lower half of that, so our equation for the cross-section becomes y = a - sqrt(a^2 - x^2).

I am considering the ball to be (momentarily) at the point (x, h) and I'm going to do a FBD about this point. So I'm going to need two things before analyzing the FBD. I need to find x, which is the radius of the circle the ball is making, and I need to know at what angle the normal force acts at this point.

The radius is easy...x = r and y = h, so from the circle equation:
y = a - sqrt(a^2 - x^2)

h = a - sqrt(a^2 - r^2)

I get r = sqrt(2ah - h^2).

Now to the hard part. First of all, going back to your Calculus I class given a function y(x) we know that y' = tan(alpha), where alpha is the angle between the tangent line to the curve and the +x axis. In this case, that means
y = a - sqrt(a^2 - x^2)

y' = -x/sqrt(a^2 - x^2) = -x/(y - a)

Putting in the point (x, y) = (r, h) gives

y' = -r/(h - a) = r/(a - h) = tan(alpha)

(We'll need the tangent in a moment.)

Now that we have the angle we can move onto the FBD. I have a weight acting downward (-y direction) and a normal force acting perpendicular to the tangent line at the point (x, y) = (r, h). Notice that the ball moves in a circle, so the net force in the x direction is equal to the centripetal force. So we get
(sum)Fx = -N*sin(alpha) = Fc = -mv^2 / r

In the y direction we get
(sum)Fy = N*cos(alpha) - mg = 0

N = mg/cos(alpha)

Putting this N into the (sum)Fx equation:
mv^2 / r = mg/cos(alpha) * sin(alpha) = mg*tan(alpha)

v^2 = gr*sqrt(2ah - h^2)/(a - h)

And since r = sqrt(2ah - h^2)

v^2 = g*sqrt(2ah - h^2)*sqrt(2ah - h^2)/(a - h)

which you can finish simplifying yourself.

-Dan
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Old Mar 13th 2011, 09:24 PM   #3
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Thank you kindly for your help, i apologise for the lateness of my response. I am reviewing your post and examining the question and solution you have given. I will make sure i understand exactly what is happening in the solution you have given.
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Old Mar 14th 2011, 09:54 AM   #4
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Originally Posted by GrahamJamesK View Post
Thank you kindly for your help, i apologise for the lateness of my response. I am reviewing your post and examining the question and solution you have given. I will make sure i understand exactly what is happening in the solution you have given.
That's all anyone can ask for. Please feel free to post if there is something you don't understand.

-Dan
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Old Mar 14th 2011, 02:18 PM   #5
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"So. Let's assume our spherical bowl has its minimum at the origin (with the usual +x and +y directions). I am going to arbitrarily pick the problem to be in the xy plane...any cross-section of the problem will do. Our segment of the sphere in the xy plane is given by x^2 + (y - a)^2 = a^2. I am choosing to work in the lower half of that, so our equation for the cross-section becomes y = a - sqrt(a^2 - x^2)."

I don't understand how the equation for the segment of the sphere has been derived? Is it based upon this equation for the radius of a sphere
r=(h2+r_b^2)/(2h) or something similar?

Thanks
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Old Mar 14th 2011, 02:56 PM   #6
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Originally Posted by GrahamJamesK View Post
"So. Let's assume our spherical bowl has its minimum at the origin (with the usual +x and +y directions). I am going to arbitrarily pick the problem to be in the xy plane...any cross-section of the problem will do. Our segment of the sphere in the xy plane is given by x^2 + (y - a)^2 = a^2. I am choosing to work in the lower half of that, so our equation for the cross-section becomes y = a - sqrt(a^2 - x^2)."

I don't understand how the equation for the segment of the sphere has been derived? Is it based upon this equation for the radius of a sphere
r=(h2+r_b^2)/(2h) or something similar?

Thanks
The most general form of the equation of a circle in the xy plane is
(x - h)^2 + (y - k)^2 = R^2

for a center (h, k) and radius R. Since the radius of our circle is a, and we know that it just touches the origin, (and that the center is on the y axis) we can state that the center of the circle is (0, a). Thus

x^2 + (y - a)^2 = a^2

-Dan
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Last edited by topsquark; Mar 24th 2011 at 10:12 AM.
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Old Mar 22nd 2011, 08:49 PM   #7
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Question

Ok thanks Dan, I understand the source of the equation now. There are a couple of things that are still confusing me before i move further into the solution.

If (x - h)^2 + (y - k)^2 = r^2; where h = 0, k = a, and a = d (the diameter) = 2r

Should the equation look like this; (i'll do it in detail so any mistakes can be easily identified)

(x-0)^2 + (y - a)^2 = (a/2)^2; (to give r^2)

x^2 + (y - a)^2 = (a/2)^2

Then isolating y:

add x^2

(y - a)^2 = (a/2)^2 - x^2

take sqrt

y - a= sqrt[(a/2)^2 - x^2]

add a

y = a + sqrt[(a/2)^2 - x^2]

Where have I gone wrong here to arrive at this conclusion? I'm sorry if this is painstakingly slow. I can't move on through the solution until I understand what I am trying to show in the solution, what tools I need to use.... and why!!!

Thanks for your patience
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Old Mar 23rd 2011, 05:54 PM   #8
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Originally Posted by GrahamJamesK View Post
Ok thanks Dan, I understand the source of the equation now. There are a couple of things that are still confusing me before i move further into the solution.

If (x - h)^2 + (y - k)^2 = r^2; where h = 0, k = a, and a = d (the diameter) = 2r

Should the equation look like this; (i'll do it in detail so any mistakes can be easily identified)

(x-0)^2 + (y - a)^2 = (a/2)^2; (to give r^2)
a is the radius, not the diameter. So
(x - 0)^2 + (y - a)^2 = a^2

Otherwise it looks good.

-Dan
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Old Mar 23rd 2011, 08:43 PM   #9
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OK, some of the labels are confusing. I was basing that on an earlier post where you said the diameter was a, not to be picking at straws, just to illustrate where i got it from. I'm glad this is starting to sink in at last!!!!Cheers

Graham

Last edited by GrahamJamesK; Mar 23rd 2011 at 08:57 PM.
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Old Mar 24th 2011, 10:13 AM   #10
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Originally Posted by GrahamJamesK View Post
OK, some of the labels are confusing. I was basing that on an earlier post where you said the diameter was a, not to be picking at straws, just to illustrate where i got it from. I'm glad this is starting to sink in at last!!!!Cheers

Graham
It seems that I did. It was supposed to be radius. I have fixed it in the post. Thanks for the catch.

-Dan
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