What Forces Act on a Pulley System with Two Masses?

[Chestermiller]

I don't seem to be having as much fun .

What was the mistake in the earlier analysis where we found right mass going down and left moving up ?
If origin is placed at the pulley and downwards is considered positive , shouldn't that be
$$mg-T=m\left[-\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2\right]$$

No. If there were no horizontal velocity, the downward acceleration should be $d^2r/dt^2$, not $-d^2r/dt^2$.

 

[Jahnavi]

OK. If we put in the subscripts, your equation for the right mass at $t = 0$ is $$mg-T=m\left[-\frac{d^2r_1}{dt^2}-r_2\left(\frac{d\theta}{dt}\right)^2\right]$$ or $$mg-T=m\left[+\frac{d^2r_2}{dt^2}-r_2\left(\frac{d\theta}{dt}\right)^2\right]$$

For the left mass, $$mg -T =m\frac{d^2r_1}{dt^2} = -m\frac{d^2r_2}{dt^2}$$

Do the signs look right? If so, solve these for $\frac{d^2r_2}{dt^2}$.

$\frac{d^2r_2}{dt^2} = +\frac{r_2}{2}\left(\frac{d\theta}{dt}\right)^2$

What does positive sign signify ? Isn't $r_2$ also the displacement of right mass as measured from origin i.e pulley ? Doesn't this tell us that right mass has positive acceleration downwards ?

 

[TSny]

$\frac{d^2r_2}{dt^2} = +\frac{r_2}{2}\left(\frac{d\theta}{dt}\right)^2$

OK. We can write this as $\frac{d^2r_2}{dt^2} = +\frac{v_0^2}{2l_0}$, where $l_0$ is the initial length of the string on each side.

What does positive sign signify ? Isn't $r_2$ also the displacement of right mass as measured from origin i.e pulley ?

Yes.

Doesn't this tell us that right mass has positive acceleration downwards ?

No. $\frac{d^2r_2}{dt^2}$ is not the vertical acceleration. The vertical acceleration at $t=0$ also includes a contribution from centripetal acceleration. See second equation in post #39.

 

[Jahnavi]

OK . Vertical acceleration is y = rcosθ and

At time zero, this is:
$$\frac{d^2y}{dt^2}=\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2$$

This gives

$$\frac{d^2y}{dt^2}=-\frac{v_0^2}{2l}$$ . Minus sign signifies that right mass has vertical acceleration in upward direction .

Similarly for left mass ,$y= r_1$

$\frac{d^2r_1}{dt^2} = -\frac{d^2r_2}{dt^2}= -\frac{v_0^2}{2l}$

$\frac{d^2y_1}{dt^2} = -\frac{v_0^2}{2l}$

Left mass also accelerates upwards

Both left and right masses accelerate upwards with same magnitude .

 

[TSny]

Or you can go back to
$$mg-T=m\left[+\frac{d^2r_2}{dt^2}-r_2\left(\frac{d\theta}{dt}\right)^2\right]$$

The left side is the net force in the downward direction. So, the expression in the brackets on the right is the downward acceleration $\frac{d^2y}{dt^2}$ if positive y is down. The quantity in the brackets can be negative even when $\frac{d^2r_2}{dt^2}$ is positive.

 

[TSny]

OK . Vertical acceleration is y = rcosθ and

At time zero, this is:
$$\frac{d^2y}{dt^2}=\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2$$

This gives

$$\frac{d^2y}{dt^2}=-\frac{v_0^2}{2l}$$ . Minus sign signifies that right mass has vertical acceleration in upward direction .

Similarly for left mass ,$y= r_1$

$\frac{d^2r_1}{dt^2} = -\frac{d^2r_2}{dt^2}= -\frac{v_0^2}{2l}$

$\frac{d^2y_1}{dt^2} = -\frac{v_0^2}{2l}$

Left mass also accelerates upwards

Both left and right masses accelerate upwards with same magnitude .

Yes. Good. This is only at t = 0.

 

[Jahnavi]

Or you can go back to
$$mg-T=m\left[+\frac{d^2r_2}{dt^2}-r_2\left(\frac{d\theta}{dt}\right)^2\right]$$

The left side is the net force in the downward direction. So, the expression in the brackets on the right is the downward acceleration $\frac{d^2y}{dt^2}$ if positive y is down. The quantity in the brackets can be negative even when $\frac{d^2r_2}{dt^2}$ is positive.

There are two set of brackets . Which one are you referring ?

 

[TSny]

There are two set of brackets . Which one are you referring ?

The square brackets [ ]. The acceleration is the coefficient of m.

 

[haruspex]

There are two set of brackets . Which one are you referring ?

Strictly speaking, () are parentheses, [] are brackets, {} are braces.

 

[Jahnavi]

Right .

How did you infer that upward acceleration of left mass will be more than the right mass after some time ?

Initially they both have the same magnitude of vertical accelerations .

 

[TSny]

Your original approach was good. You noted that the downward force on m₂ is greater than the downward force on m₁ (except initially when θ = 0 and the forces are equal). Equivalently, you could say that the upward force on m₂ is less than the upward force on m₁. So, the upward acceleration of m₂ is less than the upward acceleration of m₁. Thus m₁ rises more quickly.

 

[Jahnavi]

Your original approach was good. You noted that the downward force on m₂ is greater than the downward force on m₁ (except initially when θ = 0 and the forces are equal). Equivalently, you could say that the upward force on m₂ is less than the upward force on m₁. So, the upward acceleration of m₂ is less than the upward acceleration of m₁. Thus m₁ rises more quickly.

What about this reasoning ?

After a very short time , right mass would be moving horizontally towards right .This would lead to lengthening of right string .Correspondingly left string shortens and in the process left mass moves upwards .Left mass would be higher than right mass .

 

[Jahnavi]

Yes, I enjoyed this one a lot.

Now , I can also say , ME TOO

 

[TSny]

What about this reasoning ?

After a very short time , right mass would be moving horizontally towards right .This would lead to lengthening of right string .Correspondingly left string shortens and in the process left mass moves upwards .Left mass would be higher than right mass .

After a very short but finite time, the right mass would be moving both to the right and upward.
At t = 0, the right mass is only moving to the right. But at this instant the string on the right is not getting longer ($\dot r_2 = 0$). Neither mass is moving upward at t = 0. So, the argument is a little vague. But the argument is suggestive of what's going on.

 

[TSny]

Now , I can also say , ME TOO

That is very good to hear

 

[Jahnavi]

I learned a lot from you TSny . Excellent discussion !

Thanks a lot !

A big Thank You @Chestermiller