If a cylinder 5cm diameter and 12cm high is on a train, how fast does train need to deceletate to tip it over?

Question

yes it is a cylinder and it jsut needs to tip over

Answer 1

A very nice question.

There are two forces operating on the cylinder: gravity and the frictional force.

When the cylinder is at rest, the frictional force is zero (horizontal acceleration is 0) and the force of gravity acts over the entire end surface.

But as the acceleration increases, the frictional force generates a torque about the center of gravity. If the cylinder is not to tip over, the force of gravity must shift from being evenly distributed to being concentrated where it can counter-act the torque generated by the frictional force.

Just before the cylinder tips over, all the support is at one point at the very edge of the end of the cylinder. At this point, gravity is acting with an arm equal to the radius, while the frictional force is acting with an arm equal to half the length.

At this point, the two torques just balance. You know the acceleration due to gravity and its arm length, therefor you know its torque (in terms of M, the mass of the cylinder). You also know the arm for the frictional force and so can figure out the corresponding acceleration.

Answer 2

a/g = 5/12
a = 9.81∙(5/12)
a = 4.1 m/s^2

To visualize the solution, draw a big can, 12cm by 5cm. Draw an arrow from the center of the can pointing down, 9.81cm long. From the tip of that arrow, draw another 4.1cm long arrow to the left [or right, depending on which way the train is accelerating]. The resultant of the two arrows shows a can on the brink of toppling.