A plank with a mass M =6 kg rides on top of tweo identical solid cyindrical rollers?

Question

The rollers have a Radius R = 5cm and a mass m = 2 kg. The plank is pulled by a constant horizontal force F of magnitude 6 N applied to one end of the plank and perpendicular to the axes of the cylinders (Which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank

a) What is the acceleration of the plank and the rollers.

b) what friction forces are acting

Answer 1

A nice little problem.

There are a number of approaches and it isn't at all clear which is best, but I'd probably take the indirect approach and use energy.

Suppose the force is applied for a distance of 1 meter. Force x Distance = energy, so this gives us a total energy budget.

This is split between:

1. the kinetic energy of the plank

2. the linear kinetic energy of the rollers

3. the rotational kinetic energy of the rollers

Let V be the velocity of the plank at the end of the 1 meter trip. The kinetic energy of the plank is given by (1/2)Mp V^2.

The velocity of the top of the rollers must match that of the plank (because there is no slippage). Similarly, the velocity of the bottom of the rollers must be 0.

So the velocity of the center of gravity of the rollers is V/2, which lets you figure out the linear kinetic energy of the rollers.

The angular velocity of the rollers must be such at at the rim, the velocity relative to the center of each roller is V/2 (velocity of the center + relative velocity at the rim = V; velocity of the center - relative velocity at the rim = 0)

Since v = wr,  the angular velocity of the rollers = v/r = (V/2)/R.

The rotational kinetic energy of the rollers is given by (1/2)I w^2. To compute it, you need to compute I, the moment of inertia. But you have the mass and the radius of the rollers and are told there are homogenous (well, "solid" but assume they are homogenous) so you can compute I:

http://en.wikipedia.org/wiki/List_of_mom...

So now you have all three energy terms in terms of V^2 so you can compute V.

The equations for constant acceleration are:

Vf = Vi + AT

D = Vi T + (1/2) A T^2

You have Vi (= 0), D (= 1), and Vf (= V) so you have two equations in two unknowns which you can solve for A and T.

As for the friction forces, the total force, Ft, on the plank equals the applied force, Fa, minus the friction force Ff.

Ft = Fa - Ff

But you also know that Ft = M A, where M is the mass of the plank alone and A is the acceleration of the plank.

Since you have Ft and Fa, you can compute the friction force on the plank, Ff, and hence the horizontal force the plank exerts on each of the rollers.

Since you know the final angular velocity of the rollers, and the time interval, you can compute the angular acceleration. With the moment of inertia, you can compute the required torque.

For each roller you have the frictional force from the plank and the frictional force from the flat surface. The two form a torque. You know the total torque and the torque due to the plank force, so you can compute the surface frictional force.