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Denote the contact points of the ball with the rods with A and B.

Let R' be radius of the circle at the same plane with the rods, and φ be the angle formed by the circle's arc and y-axis. From picture, we can get


and

From equation (1), we can get


Position vector of point A and B is



Substitute R' with R and φ (from now, we work with φ and θ),



Thus,




Normal forces at point A and B is pointing inward to the center. Assume magnitude of the normal force is N. Then vector of the forces at point A and B,



Thus,





And for the friction forces that work on the ball, are parallel to the rods. So we can easily determine the directions. Assume magnitude of both friction forces is f. Thus vector of the friction forces at point A and B,



Only friction forces that cause torque on the ball, so the total torque on the ball is,


Substitute and from equations (3) and (4), and substitute and from equations (7) and (8), resulting


And from , we can get,



And now, we need to calculate the acceleration of ball's CM.



Substitute all forces from equations (5), (6), (7), (8),




Because the ball is not slip, then the total acceleration at contact points must be zero,


Substitute and from equation (10) and (11),


From the equation above, we can get two equations by separating x component and y component,



Solving these two equations, and we get


Substitute f from equation above into equation (10), and we get


And then substitute φ from equation (2),





That's all my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Solution - V-shaped Rod PathSocialTwist Tell-a-Friend

2 comments

Yen said... @ March 25, 2009 at 2:45 AM

Hi, it's nice to find your site, i like this site.
little correction, i think you got a wrong solution on this problem, you use wrong coordinate. I investigate from your equation, your x axis pointed to the right (on the picture), but actually it should point to left, because you take y axis vertical downward and z axis point out of the screen.
I found the answer should be: mgl(sin(theta))^2/(I+m(R^2-(lsin(theta)cos(theta))^2)
and actually there is another way (easier I think) to solve it, without vectors. You can use parallel axis theorem to find the moment of inertia about axis passing through AB (these points are at rest, so it construct the instantaneous axis of rotation), and use the non vector equation torque = (moment of inertia)(angular acceleration), what nice here is, torque of the frictions and normals are zero about this axis, no need to calculate them.
Thanks, looking forward for your next problem :)

Yen said... @ March 25, 2009 at 3:38 AM

Hi again, sorry, I just realize you put your coordinate axis on the problems picture, if that's your coordinate, then your equations are right, just interchange the point A and B in the picture.
I try to solve your equations and they gave me the same answer I got. I found your answer is wrong at equation (11). Sorry if I make you confuse.
Thanks

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