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.