# Homework Help: Period of Oscillation of a Board Between 2 Identical Rollers

1. May 13, 2012

1. The problem statement, all variables and given/known data
Two identical rollers are mounted with their axes parallel, in a horizontal plane, a
distance 2d = 26.5 cm apart. The two rollers are rotating inwardly at the top
with the same angular speed (w). A long uniform board is laid across them in a direction
perpendicular to their axes. The board of mass m = 3.52 kg is originally placed so that
its center of mass lies a distance x(initial) = 10 cm from the point midway between the rollers.
The coecient of friction between the board and rollers is k = 0.653. What is the
period (s) of the motion?

2. Relevant equations
Merg, I'm not sure?
x(t) = Acos(ωt+phi)
T = 2∏(sq)(l/2μg)

3. The attempt at a solution

I did attempt it, based on the advice of my tutor and what he found on this board earlier, however, I got something...not right. Not even close. The real answer is .90.

I played around with it a bit, trying to put in the actual normal force of the board and such. I feel like the center of mass and where it starts from is important, but i can't figure out how to incorporate either.

2. May 13, 2012

### tiny-tim

welcome to pf!

what are the two normal forces as a function of x ?

and so what is the net friction force as a function of x ?

3. May 13, 2012

Hi Tiny Tim!

If you don't mind, I'm going to try to think this one out on the board. You asked what are the two normal forces as a function of x? and so what is the net friction force as a function of x ?

Ok, so I know that depending on where x is, the normal forces of the wheels against the board are going to change. As the board moves one way, the normal force and frictional force will increase on the wheel overwhich more of the weight of the board is and less on the other wheel. Eventually, the high frictional force on one wheel will drive the board in the opposite direction, toward the wheel with the lower frictional force.. However, I have *no* clue how to even begin writing an equation that reflects that.

Last edited: May 13, 2012
4. May 13, 2012

### tiny-tim

yes … so how would you find the values of the two normal forces?

5. May 13, 2012

Ʃf = μn1+ μn2

erg, sorry that's frictional. For normal force I would use
Ʃf = m1g+ m2g

Okay, so I guess I need to use the center of mass equation to figure out how much mass is over each roller?

6. May 13, 2012

### tiny-tim

yes, or take moments about one roller to find the normal force at the other roller

7. May 13, 2012

I don't know what you mean by that.

8. May 13, 2012

### tiny-tim

the board isn't rotating, so the moments of forces about any point must add to zero

9. May 13, 2012

So
0 = m1g+ m2g

10. May 13, 2012

### tiny-tim

??

do you know what "taking moments" means?

11. May 13, 2012

I thought you knew that I don't know what that means because you said, "take moments" and I said, "I dont know what you mean by that."

I don't know what you mean by "moments".

12. May 13, 2012