|Jan14-11, 10:39 AM||#1|
Sliding or toppling cylinder.
Greetings. I answer a lot of physics questions on another forum and I came across this one. So it's someone's homework problem, but not mine. :-) But as I looked at it, I realized that it's either very simple and I'm missing something, or it's more subtle and I don't have the tools to solve it. I'd appreciate knowing which. It really just intrigued me because it wasn't simple.
1. The problem statement, all variables and given/known data
A cylindrical shape with height 70cm and diameter 4cm is on a plane inclined at 30 degrees. The coefficient of friction is tan(30) (I'm assuming 30 degrees here). Will the shape slide or topple?
3. The attempt at a solution
I initially thought all I'd have to do was calculate the sum of torques about the front edge of the cylinder as it slides down. But then I realized that's insufficient because the front edge is accelerating (assuming it's sliding). Also, for a static cylinder, I'd just have a moment from the center of gravity, and I'd have a force from friction. But all the frictional force would be on the plane the same as the tipping point, so there would be no moment.
Next, although the cylinder is accelerating, I thought it's possible that it would be valid to use it as a frame of reference. In the no friction case, the cylinder will be experiencing an acceleration that is normal to the plane, so it's stable. As friction increases, the acceleration vector will shift to one side. Once it passes the tipping point, the shape will topple. In the cylinder frame of reference, the CG is 35cm up and 2cm from the edge, so the frictional acceleration can be no more than 2/35 of the normal acceleration. I could solve for the coefficient of friction that will be on the boundary.
acc(norm) = F(normal) / m
acc(norm) = m * g * cos(30) / m = g * cos(30)
acc(plane) = 2/35 * g * cos(30)
F(plane) = 2/35 * g * cos(30) * m = mu * m * g * cos30
2/35 * cos(30) = mu * cos(30)
2/35 = mu
Since that's much smaller than tan(30), it'll topple. But this doesn't feel right to me. So I figure I'm missing a bit. I well believe this method I took isn't valid, but thought I'd work through it just to see.
Thanks for helping me out!
|Jan14-11, 02:46 PM||#2|
Hi BowlOfRed! Welcome to PF!
you can use any "comoving" point, ie any point whose velocity is parallel to the velocity of the centre of mass …
in particular for a body that isn't rotating, you can use any point on the body.
(don't forget that the angular momentum isn't Iω = 0, there's an extra v term)
Alternatively, treating it as a statics problem with a fictitious m dv/dt force parallel to the slope should give the same result.
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