# Motion of bar on rollers

1. Oct 12, 2013

### Saitama

1. The problem statement, all variables and given/known data
A heavy uniform bar of mass M rests on top of two identical rollers which are continuously turned rapidly in opposite directions, as shown. The centres of rollers are a distance 2l apart. The coefficient of friction between the bar and the roller surface is $\mu$, a constant independent of the relative speed of the two surfaces.

Initially the bar is held at rest with its centre at distance $x_0$ from the midpoint of the rollers. At time t=0 it is released. Find the subsequent motion of the bar.

2. Relevant equations

3. The attempt at a solution
The forces acting on the bar are its own weight, normal reactions from the rollers and force of friction due to relative motion between the rollers and the bar. But the problem is how do I determine the direction of friction? I have this dilemma from a long time and I hope to clear it by means of this thread.

Any help is appreciated. Thanks!

#### Attached Files:

• ###### bar on rollers.png
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2. Oct 13, 2013

### Tanya Sharma

First consider the left roller.

Just focus on the surfaces in contact between the roller and the bar. If there were no friction,where would the topmost point of the roller move with respect to the bar,towards left or right ?

3. Oct 13, 2013

### Saitama

If there were no friction, bar would move to the right. Then the friction on bar would be acting towards left, right?

Similarly, for the right roller, the friction on bar would towards left, correct?

4. Oct 13, 2013

### ehild

Think of the essential property of the force of friction: that it opposes relative motion between the surfaces in contact.

ehild

5. Oct 13, 2013

### Tanya Sharma

Wrong...

Please answer this - Where would the topmost point of the left roller move ?

6. Oct 13, 2013

### Saitama

Towards right. Then the friction on the roller acts towards left?

7. Oct 13, 2013

### Tanya Sharma

Good...Where would it act on the bar ?

8. Oct 13, 2013

### Saitama

Towards right. Call it $f_1$

Similarly, for the right roller, the topmost point moves towards left so the friction on the bar ($f_2$) acts towards left.

Displace the bar by $x$ towards right.
Applying Newton's second law in horizontal direction,
$$F=f_1-f_2=\mu N_1-\mu N_2$$

In the vertical direction,
$$Mg=N_1+N_2$$

$$N_1(l+x)=N_2(l-x)$$

Are my equations correct?

#### Attached Files:

• ###### bar on rollers fbd.png
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9. Oct 13, 2013

### Tanya Sharma

Correct

Last edited: Oct 13, 2013
10. Oct 13, 2013

### Saitama

I solved the equations and ended up with this:

$$M\frac{d^2x}{dt^2}=-\frac{\mu Mgx}{l}$$

The solution to the above differential equation is:
$$x(t)=A\cos(\omega t)+B\sin(\omega t)$$
where $\omega=\sqrt{\mu g/l}$.

From the initial condition, $x(0)=x_0$ and $x'(0)=0$. Hence, $A=x_0$ and $B=0$.

$$x(t)=x_0\cos(\omega t)$$

Looks good?

11. Oct 13, 2013

### Tanya Sharma

:thumbs:

12. Oct 13, 2013

### Saitama

Thanks a lot Tanya! :)