# Homework Help: Mechanics problem: cylinder on cart

1. Oct 21, 2005

### positron

I am having problems with the following:
A uniform 2 kg cylinder rests on a laboratory cart. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in
diameter and 10 cm in height, which of the following is closest to the minimum acceleration of the cart needed to cause the cylinder to tip over?
[A] 2 m/s2 4 m/s2 [C] 5 m/s2 [D] 6 m/s2 [E] The cylinder would
slide at all of these accelerations.

This is my reasoning so far:
The moment of inertia of the cylinder about its center of mass is I = 1/4*M*R^2 + 1/12*M*L^2 = 1/4*2*0.02^2 + 1/12*2*0.10^2 = .00186. The torque about the center of mass caused by friction is F*d = mu*M*g*(sqrt(.29)) = 0.05*2*10*sqrt(.05^2 + .02^2) = 0.0539.
Since torque = I*alpha, where alpha is the angular acceleration, this gives an angular cceleration of 28.95 rad/s^2. The linear acceleration is alpha*r = 28.95 rad/s^2*sqrt(.05^2 + .02^2) m = 1.56.

You don't have the check the numbers, but please point out if there is something obviously wrong in my reasoning.
Thanks.

2. Oct 21, 2005

### lightgrav

First, you calculated the MAXIMUM Friction Force
(at zero acceleration the ACTUAL F_fr = 0).

Second, if the cart accelerates, F_fr is NOT zero,
so the c.o.m. of the cylinder accelerates (x-direction).
But if the cylinder starts to tip, the Normal Force
shifts from straight underneath the c.o.m.,
to the "back edge", with lever-arm R from c.o.m.

You need F_fr torque to be greater than (or equal)
to N's torque, both around the c.o.m.