Mechanics problem: cylinder on cart

[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.
The answer is B.)

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.

 

[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.

By the way, your reasoning about momentum & KE in collisions
was fine, but you need to start using more precise words;
"differ" and "different", "distinct", "dissimilar", "massive"...
It is often useful to write KE = ½pv or as (p^2)/(2m).

 

[quicknote]

Your reasoning appears to be correct. To determine the minimum acceleration needed to cause the cylinder to tip over, we need to consider the forces and torques acting on the cylinder. The torque caused by friction, as you calculated, is equal to the moment of inertia of the cylinder times its angular acceleration. This angular acceleration can then be converted to a linear acceleration by multiplying by the radius of the cylinder.

One suggestion for improvement would be to include units in your calculations and final answer. This helps to ensure that all the values are consistent and makes it easier to interpret the results.

Another consideration is that the minimum acceleration needed to tip over the cylinder may not be the same as the acceleration needed to cause it to slide. This is because the friction force will decrease as the cylinder starts to tip, making it easier for the cart to accelerate. So, while the answer may be B), it is possible that the cylinder may start to slide at a lower acceleration. This would require further analysis to determine.