What Forces Act on a Pulley System with Two Masses?

[TSny]

If left mass moves upwards shouldn't right move downwards ?

Not necessarily. If you move the right mass by hand so that it moves mostly horizontally but a little bit upward also, what would the first mass do?

 

[Jahnavi]

Not necessarily. If you move the right mass by hand so that it moves mostly horizontally but a little bit upward also, what would the first mass do?

Move upward .

$\frac{dV_v}{dt} = - \frac{V_h^2}{2L}$

Doesn't this tell us that right mass moves downwards ?

If not , what does this equation convey ?

 

[Chestermiller]

If $y_2$ is the height above the floor of the swinging mass, then does the symbol $V_v$ equal $\dot y_2$? If so, then initially wouldn't we have $$T-mg= m \ddot y_2 = m\frac{dV_v}{dt} \,\,\,\,?$$

No. There is a term missing in the acceleration. If r is the distance of the right mass from the pulley, then, using cylindrical coordinates, the radial acceleration of the mass is:
$$-\omega^2 r+\frac{d^2r}{dt^2}$$
where $$\omega=\frac{V_h}{r}$$and$$\frac{d^2r}{dt^2}=-\frac{d^2y_2}{dt^2}=-\frac{dV_v}{dt}$$
So, at time zero, the upward acceleration is $$\frac{V_h^2}{r}+\frac{dV_v}{dt}$$
The missing term is the centripetal acceleration.

 

[Chestermiller]

Move upward .Doesn't this tell us that right mass moves downwards ?

If not , what does this equation convey ?

Yes, it tells us that, initially, the length of the right section of string is increasing, and the length of the left section of string is decreasing.

 

[TSny]

No. There is a term missing in the acceleration. If r is the distance of the right mass from the pulley, then, using cylindrical coordinates, the radial acceleration of the mass is:
$$-\omega^2 r+\frac{d^2r}{dt^2}$$
where $$\omega=\frac{V_h}{r}$$and$$\frac{d^2r}{dt^2}=-\frac{d^2y_2}{dt^2}=-\frac{dV_v}{dt}$$

OK, except for the last equation.
How do you get $\frac{d^2r}{dt^2}=-\frac{d^2y_2}{dt^2}$?
Imagine that the second mass is contrained to move in a horizontal track. Then $\frac{d^2y_2}{dt^2} = 0$, but $\frac{d^2r}{dt^2}$ would not need to be zero.

 

[TSny]

Doesn't this tell us that right mass moves downwards ?

I don't think that the equation $\frac{dV_v}{dt} = - \frac{v_h^2}{2L}$ is correct. I believe a correct analysis will give $\frac{dV_v}{dt} = + \frac{v_h^2}{2L}$.

 

[Chestermiller]

OK, except for the last equation.
How do you get $\frac{d^2r}{dt^2}=-\frac{d^2y_2}{dt^2}$?
Imagine that the second mass is contrained to move in a horizontal track. Then $\frac{d^2y_2}{dt^2} = 0$, but $\frac{d^2r}{dt^2}$ would not need to be zero.

It's not going to be moving on a horizontal track. To me it's very clear that, at t = 0, $\frac{d^2r}{dt^2}=-\frac{d^2y_2}{dt^2}$. Even though the vertical velocity is zero, the rate of change of vertical velocity is not zero.

 

[Chestermiller]

I don't think that the equation $\frac{dV_v}{dt} = - \frac{v_h^2}{2L}$ is correct. I believe a correct analysis will give $\frac{dV_v}{dt} = + \frac{v_h^2}{2L}$.

I don't think I made a sign error in my analysis (especially using cylindrical coordinates), but maybe I did. If you can spot a sign error, I would be pleased if you could point it out. But, to me, the analysis clearly shows that the left mass will be accelerating upward.

 

[Chestermiller]

Second thoughts. I stand corrected.

OK, if I write $y=r\cos{\theta}$ (representing the vertical distance from the elevation of the right mass to the elevation of the pulley), then the downward acceleration of the right mass is $$\frac{d^2y}{dt^2}=\cos{\theta}\frac{d^2r}{dt^2}-2\sin{\theta}\frac{dr}{dt}\frac{d\theta}{dt}-\cos{\theta}r\left(\frac{d\theta}{dt}\right)^2$$At time zero, this is:
$$\frac{d^2y}{dt^2}=\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2$$

Thanks for pointing this all out. So, for the right mass, $$T-mg=m\left[-\frac{d^2r}{dt^2}+r\left(\frac{d\theta}{dt}\right)^2\right]$$For the left mass, $$T-mg=m\frac{d^2r}{dt^2}$$So, $$\frac{d^2r}{dt^2}=\frac{1}{2}\frac{V_h^2}{r}$$
So, the left mass is accelerating upward.

 

[TSny]

$$\frac{d^2r}{dt^2}=\frac{1}{2}\frac{V_h^2}{r}$$
So, the left mass is accelerating upward.

Yes. Initially, both masses experience the same upward force T - Mg and both masses have the same upward acceleration $\frac{1}{2}\frac{V_h^2}{r}$ at this instant.

 

[Chestermiller]

Yes. Initially, both masses experience the same upward force T - Mg and both masses have the same upward acceleration $\frac{1}{2}\frac{V_h^2}{r}$ at this instant.

Yes! So, considering the wording of the original problem statement, where does this leave us? Certainly, at very short times, both masses will rise vertically the same amount. Do you interpret this as meaning that answer C is correct?

 

[TSny]

So, considering the wording of the original problem statement, where does this leave us? Certainly, at very short times, both masses will rise vertically the same amount. Do you interpret this as meaning that answer C is correct?

I'm leaning more towards B. I interpret "after some time" to be more than a very small time. Technically, the mass on the left will always be higher than the mass on the right for any finite time t > 0.

 

[Chestermiller]

I'm leaning more towards B. I interpret "after some time" to be more than a very small time. Technically, the mass on the left will always be higher than the mass on the right for any finite time t > 0.

I agree. Certainly the length of string on the left will be shorter than the length of string on the right.

 

[TSny]

I agree. Certainly the length of string on the left will be shorter than the length of string on the right.

Yes. I used Mathematica to get a numerical solution to the equations of motion. I let the initial length on both sides be 1 m and let the initial speed of m₂ be 2.5 m/s. Here is the trajectory of m₂ for the first 1.5 seconds. The origin is at the initial position of m₂. You can see how m₂ rises upward before starting its descent.

Here are plots of just the y-coordinate of each mass as a function of time

Here's a closer look at the y-coordinates near the initial time. For about the first tenth of a second, the y coordinates are almost the same. But then you can see how y₂ falls behind y₁.

 

[Chestermiller]

Yes. I used Mathematica to get a numerical solution to the equations of motion. I let the initial length on both sides be 1 m and let the initial speed of m₂ be 2.5 m/s. Here is the trajectory of m₂ for the first 1.5 seconds. The origin is at the initial position of m₂. You can see how m₂ rises upward before starting its descent.
https://www.physicsforums.com/attachments/210591Here are plots of just the y-coordinate of each mass as a function of time
View attachment 210592

Here's a closer look at the y-coordinates near the initial time. For about the first tenth of a second, the y coordinates are almost the same. But then you can see how y₂ falls behind y₁.
View attachment 210593

I love it! Your modeling analysis was spot in.

This has been a really fun problem to work on.

 

[TSny]

I love it! Your modeling analysis was spot in.

This has been a really fun problem to work on.

Thanks. Yes, I enjoyed this one a lot.

 

[Jahnavi]

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 ?

So, for the right mass, $$T-mg=m\left[-\frac{d^2r}{dt^2}+r\left(\frac{d\theta}{dt}\right)^2\right]$$For the left mass, $$T-mg=m\frac{d^2r}{dt^2}$$

If origin is placed at the pulley and downwards is considered positive , shouldn't that be
$$mg-T=m\left[-\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2\right]$$ and

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

gives $\frac{dV_v}{dt} = - \frac{v_h^2}{2L}$ , which means left mass accelerates upwards.

Here , $\frac{dV_v}{dt} = \frac{d^2r}{dt^2}$

Acceleration of right mass is $-\frac{d^2r}{dt^2} = +\frac{v_h^2}{2L}$ which means it moves downwards .

What is the mistake ?

Edit : Another problem I see is that if we represent acceleration of left mass to be $\frac{d^2r}{dt^2}$ , then since length of complete string is constant , acceleration of right mass would be $- \frac{d^2r}{dt^2}$ .This way they will always have opposite signs ,hence opposite directions .If left moves up , right moves down .

 

[TSny]

Jahnavi, may I ask you to check over all of the signs in your previous post? If $r_1$ is the length of string on the left and $r_2$ the length on the right, then you are correct that $\ddot r_1 = -\ddot r_2$. I'm not sure if the $r$ that you used in your previous post is $r_1$ or $r_2$.

After we make sure all the signs are correct, then we can consider the vertical components of acceleration of the masses.

It's possible, at an instant of time, for the string on the right to be getting longer while at the same instant the mass on the right has a y-component of velocity that is upward. Similarly, it's possible for both $\ddot r_2$ and $\ddot y_2$ to be positive at the same instant (where the positive $y$-direction is upward).

 

[Jahnavi]

I'm not sure if the $r$ that you are used in your previous post is $r_1$ or $r_2$.

$r$ is the distance of left mass from the pulley .Since left string length is $r_1$ , I suppose $r = r_1$ .

Since $\ddot{r_2}= - \ddot{r_1}$ , for the equations of the right mass I used $-\ddot{r}$ to represent $\ddot{r_2}$ .

I am assuming a common variable $r$ to represent displacements of masses as well as length of left string. Is it wrong ?

It seems there is not only confusion in my understanding but confusion in notations as well .

If it is still unclear , you can label the variables as per your understanding and I will show you the respective equations .This way there will be minimum confusion .

Surely there are problems with signs and notations .

 

[TSny]

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 $\frac{d^2r_2}{dt^2}$.

We can then go on to see why $\frac{dV_v}{dt} \neq \frac{d^2r_2}{dt^2}$. Or just look at the first couple of equations in @Chestermiller post 39.

 

[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 .