What Forces Act on a Pulley System with Two Masses?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
When the right mass moves horizontally towards right , the verticle force acting on it is Mg - Tcosϑ , where ϑ is the angle which right string makes with verticle . The left mass experiences a force Mg - T . Since verticle force experienced by right mass is more , it moves closer to ground .Hence B) should be correct .

Is that right ?

 

[BvU]

What do your relevant equations have to say about that ?

 

[Jahnavi]

What do your relevant equations have to say about that ?

Motion of masses in vertical direction are governed by respective net forces acting on them in vertical direction .The corresponding net forces have been mentioned in the post .

Is there anything more you would like me to explain ?

 

[BvU]

Yes. I don'see the speed that is imparted to the mass appearing in your reasoning. So if the speed is zero, does it stilll apply ?
And you don't say what T is either ...

Did you try to do the experiment ?

 

[Jahnavi]

And you don't say what T is either ...

Tension in the string connecting the two masses .

I don'see the speed that is imparted to the mass appearing in your reasoning. So if the speed is zero, does it stilll apply ?

Why would speed appear in force equations ?

We may consider any arbitrary horizontal speed being imparted to the right mass .

 

[Chestermiller]

Tension in the string connecting the two masses .Why would speed appear in force equations ?

We may consider any arbitrary horizontal speed being imparted to the right mass .

Please write down your force balance in the horizontal direction for the mass on the right.

 

[Jahnavi]

Please write down your force balance in the horizontal direction for the mass on the right.

Tsinθ = Ma_x

T is tension in the string and θ is angle which right string makes with vertical .

 

[Chestermiller]

At the instant after the mass on the right has received its horizontal velocity, what is the angle theta, and what is its centripetal acceleration with respect to the point of contact of the string with the pulley?

 

[Jahnavi]

what is the angle theta

zero .

what is its centripetal acceleration with respect to the point of contact of the string with the pulley?

v²/L

L is half length of string .

 

[Chestermiller]

zero .v²/L

L is half length of string .

OK. So, initially the vertical force balance on the right mass should really be: $$T-mg=m\frac{V_h^2}{L}+m\frac{dV_v}{dt}$$where $V_h$ is the initial horizontal velocity of the right mass and $\frac{dV_v}{dt}$ is the rate of change of its vertical velocity component.

 

[Jahnavi]

Ok . What should be the next step in the analysis ?

 

[Chestermiller]

Ok . What should be the next step in the analysis ?

Write the force balance on the left mass.

 

[Jahnavi]

Write the force balance on the left mass.

T - Mg = Ma_y

 

[Chestermiller]

T - Mg = Ma_y

So, what do you get if you subtract the two equations, recognizing that, for the left mass, $a_y$ is the same as $-dV_v/dt$ for the right mass?

 

[Jahnavi]

So, what do you get if you subtract the two equations, recognizing that, for the left mass, $a_y$ is the same as $-dV_v/dt$ for the right mass?

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

 

[Chestermiller]

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

Good. So what does this tell you about the initial vertical movement of the right- and left masses?

 

[Jahnavi]

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

Lengthening and shortening of a string is given by the expression $\frac{dV_v}{dt}$ .Isn't it same for both the masses ?

 

[Chestermiller]

Lengthening and shortening of a string is given by the expression $\frac{dV_v}{dt}$ .Isn't it same for both the masses ?

No. They are moving in opposite directions. The total length of string is constant, so the length of string on the right gets longer, and the length of string on the left gets shorter.

 

[Bestfrog]

Can I ask you where did you take this problem?

 

[Jahnavi]

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

Good. So what does this tell you about the initial vertical movement of the right- and left masses?

Since you took positive upwards and from the above expression of vertical acceleration of right mass has a negative sign , it's initial acceleration is downwards .Likewise acceleration of left mass is positive means it's acceleration is upwards .

Right ?

 

[Chestermiller]

Since you took positive upwards and from the above expression of vertical acceleration of right mass has a negative sign , it's initial acceleration is downwards .Likewise acceleration of left mass is positive means it's acceleration is upwards .

Right ?

right.

 

[Jahnavi]

Can I ask you where did you take this problem?

Don't know the source .It was given in one of the Mechanics Practise sheets.

 

[Jahnavi]

right.

Thank you so much . Your analysis was amazing .

 

[Jahnavi]

So , basically the left mass moves upwards and the right mass moves downwards and rightwards simultaneously .

Right ?

 

[Jahnavi]

Could you point out the exact flaw in my reasoning in first post ?

In light of your analysis , even though my initial attempt looks odd , I can't clearly see the exact problem with that reasoning .

 

[TSny]

Could you point out the exact flaw in my reasoning in first post ?

I think the basic approach in your first post is good. However, you seem to assume that Mg - T is positive so that both masses will initially have a downward component of acceleration. Is that possible if the string is not allowed to stretch?

 

[Jahnavi]

Good point .

But isn't it interesting that net force on the right mass is upwards , yet it moves downwards ?

Mathematically , it is fine , but looking from an F = Ma perspective , it is quite puzzling .

Is it because the acceleration has two components and net acceleration for right mass is still upwards (despite moving downwards ) ?

 

[TSny]

OK. So, initially the vertical force balance on the right mass should really be: $$T-mg=m\frac{V_h^2}{L}+m\frac{dV_v}{dt}$$where $V_h$ is the initial horizontal velocity of the right mass and $\frac{dV_v}{dt}$ is the rate of change of its vertical velocity component.

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 $$T-mg= m \ddot y_2 = m\frac{dV_v}{dt} \,\,\,\,?$$

 

[TSny]

Good point .

But isn't it interesting that net force on the right mass is upwards , yet it moves downwards ?

Does it really move downwards initially?

 

[Jahnavi]

It seems I am still unsure of how things are moving .

I thought right mass moves rightwards and downwards .

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

 

[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.