# Plank being pulled across two cylinders

1. Nov 11, 2012

### bvschaefer

1. The problem statement, all variables and given/known data

A plank with a mass M = 6.30 kg rides on top of two identical, solid, cylindrical rollers that have R = 5.30 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force Farrowbold of magnitude 5.00 N applied to the 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) Find the initial acceleration of the plank when the rollers are equidistant from edge of the plank

(b)Find the acceleration of the rollers at this moment

(c)What friction forces are acting at this moment? (Fg and Fp)

2. Relevant equations

F = ma
T = Iα
T = F x R

3. The attempt at a solution
Using parallel axis theorem, I have:
I = Icm + MR2
I = .5(2.00kg)(.053)2 + (2.00kg)(.053)2
I = 0.00841
where point of rotation is at the ground.

From here, I know I have to use a torque equation to solve for acceleration but am stuck, any help is appreciated!

2. Nov 11, 2012

### tiny-tim

welcome to pf!

hi bvschaefer! welcome to pf!
yes, you need the torque of F about the bottom point …

what is the difficulty with that?

(btw, you will also need an equation relating α and a)

3. Nov 11, 2012

### haruspex

4. Nov 12, 2012

### ehild

Consider the motion of the plank and the cylinder separately: The plank performs translation, the cylinders roll, which is translation of their centre of mass and rotation about the centre of mass.

Write up Newton's second low for the translational motions, collecting the forces exerted both on the plank and on the cylinders.

The friction Fp between the plank and cylinders point in opposite direction as the pulling force in case of the plank, but the friction of magnitude Fp drives the cylinders forward. The rotational resistances between the ground and the cylinder, (Fg) points also forwards.
Both forces of friction exert some torque on the cylinders

The cylinders roll without slipping that means the translational speed of their CM is equal to ωR, the acceleration of the CM is aCM=βR, (β is the angular acceleration of the forward rotation).
The plank does not slip on the cylinders: so its velocity is the same as the linear velocity of the topmost point of the cylinder. What is it compared to the velocity of the CM?

You have two equation for the cylinders: one for the translation of the CM, and one for the rotation. What are they?

ehild

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Last edited: Nov 12, 2012