# Board on top of two cylinders - inclined plane

1. Mar 21, 2013

### WannabeNewton

1. The problem statement, all variables and given/known data
See image: http://s18.postimage.org/6ql2zlo2x/board_n_cylinders.png

3. The attempt at a solution
I know it says conservation of energy but I just wanted to do it in terms of torque and forces first. The reason I'm not sure about my answer is because I barely had to use more than 3 torque + force equations so it is disconcerting...

Anyways, I used a coordinate system with the x axis along the incline and y axis perpendicular to the incline as usual with positive being downwards and outwards. There is friction between the board and each of the two cylinders at their contact point opposing forward motion of the board, pointing along the x axis of this coordinate system. Call $f_2$ the friction between the first cylinder and board and $f_3$ the friction between second cylinder and board. Note that in this coordinate system, the only acceleration is along the x - axis; call $x_1,x_2,x_3$ the x coordinates of the board, and two cylinders respectively. Hence the equation of motion for the board is $m\ddot{x_1} = mg\sin\theta - f_2 - f_3$ (note that the backwards pointing friction forces on the board from the two cylinders result in a 3rd law forward pointing friction force pair on each of the two cylinders).

Now for the two cylinders, their torque equations (with the respective reference points being from the contact points between the cylinders and the ground) are $-(\frac{1}{2}\frac{m}{2}R^{2} + \frac{m}{2}R^{2})\frac{\ddot{x_2}}{R} = -\frac{3}{4}m\ddot{x_2}R = -\frac{m}{2}gR\sin\theta - 2Rf_2$ and similarly $-\frac{3}{4}m\ddot{x_3}R = -\frac{m}{2}gR\sin\theta - 2Rf_3$ (I added the negative signs in front of the $I\alpha$ terms because the cylinders are rolling forward along the incline hence they have negative angular acceleration). At this point, we note that $\ddot{x_1} = 2\ddot{x_2} = 2\ddot{x_3}$ in order to have no slipping between them. Thus adding together the two torque equations and applying this constraint we have that $\frac{3}{4}m\ddot{x_1} = mg\sin\theta + 2f_2 + 2f_3$. Combining this with the equation of motion for the board gives $\frac{11}{4}m\ddot{x_1} = 3mg\sin\theta$ i.e. $\ddot{x_1} = \frac{12}{11}g\sin\theta$.

I'm not entirely sure this is correct because as I said I only needed one force equation , 2 torque equations, and the constraint with no need for the force equations for the cylinder nor the vanishing torque equation for the board; I would like to think it is because of the highly constrained nature of the system as given in the problem but yeah idk. If it however is fine then I would like to show my attempt at the energy / lagrangian method here as well. Thanks in advance!

2. Mar 21, 2013

### ehild

It looks correct.

ehild

3. Oct 12, 2017

### Arnav jamale

Looks correct