Motion of Bar on Rollers: Solving for Direction of Friction

[Saitama]

Homework Statement

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.

Homework Equations

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!

 

[Tanya Sharma]

Homework Statement

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.

Homework Equations

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!

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 ?

 

[Saitama]

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 ?

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?

 

[ehild]

Homework Statement

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.

Homework Equations

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?

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

ehild

 

[Tanya Sharma]

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?

Wrong...

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

 

[Saitama]

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

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

 

[Tanya Sharma]

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

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

 

[Saitama]

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

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,
$$Mg=N_1+N_2$$

Balancing moments about the CM,
$$N_1(l+x)=N_2(l-x)$$

Are my equations correct?

 

[Tanya Sharma]

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,
$$Mg=N_1+N_2$$

Balancing moments about the CM,
$$N_1(l+x)=N_2(l-x)$$

Are my equations correct?

Correct

 

[Saitama]

Correct

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 $B=0$.

So the final answer is:
$$x(t)=x_0\cos(\omega t)$$

Looks good?

 

[Tanya Sharma]

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 $B=0$.

So the final answer is:
$$x(t)=x_0\cos(\omega t)$$

Looks good?

:thumbs:

 

[Saitama]

:thumbs:

Thanks a lot Tanya! :)