Very Basic Mechanics: Equal Masses on a Pulley

[erisedk]

Homework Statement

Two equal masses hang on either side of a pulley at the same height from the ground. The mass on the right is given a horizontal speed, then after some time--
(A) The mass on the left will be nearer to ground.
(B) The mass on the right will be nearer to ground.
(C) Both the masses will be at equal distance from the ground.
(D) Nothing can be said regarding their positions.

Homework Equations

The Attempt at a Solution

Just thinking visually, I feel that as the mass on the right moves with a horizontal velocity, the left mass will lift up.
So, answer is B, which is in fact the right answer.
But the masses also feel tension and I just think there must be a more concrete explanation.

 

[gleem]

Both masses have a net force of zero i.e. the tension = weight.for each a condition of equilibrium and thus there is no acceleration but given a shove ( or impulse) they will move at constant speed depending on the magnitude of that impulse. The masses are out of equilibrium only during the time of the impulse.

 

[physichu]

look at the scetch

Let's say that the length of the swinging part of the rope is $r$ and we give the right mass a ${v_0}$ speed. Then when it's in the $\theta$ angle position. it will have a speed $v$ such that

$mgr\left( {1 - \cos \theta } \right) + {1 \over 2}mv_0^2 = {1 \over 2}m{v^2}$.

The radial forces on the right body will give

$T - \cos \theta mg = m{{{v^2}} \over r}$.

Now you need to show that $T > mg$

 

[physichu]

OOPSS.

The first equation should be

${1 \over 2}mv_0^2 = {1 \over 2}m{v^2} + mgr\left( {1 - \cos \theta } \right)$

 

[physichu]

After that corection I will say that it's depand on ${v_0}$.

 

[haruspex]

after some time

I'm not sure how to interpret that. Does it mean, for all time after some point?
If so, consider what the eventual motion will look like.

Now you need to show that T>mg

That would certainly show that the left hand mass rises initially, but it does not directly follow that it will at some point be higher than the right hand mass. When a pendulum is at one extreme of its swing, the tension is less than mg.

 

[physichu]

Not necessarily, it's depands on ${v_o}$.

 

[haruspex]

Not necessarily, it's depands on ${v_o}$.

What depends on v₀? That the left hand mass rises initially? That at some point it will be higher than the right hand mass? Whether the tension in a pendulum is less than mg at some point in its swing?

 

[physichu]

Whether the tension in a pendulum is less than mg at some point in its swing.

 

[haruspex]

Whether the tension in a pendulum is less than mg at some point in its swing.

Consider a pendulum at one extreme of its swing.

 

[erisedk]

look at the scetch
View attachment 85812

Let's say that the length of the swinging part of the rope is $r$ and we give the right mass a ${v_0}$ speed. Then when it's in the $\theta$ angle position. it will have a speed $v$ such that

$mgr\left( {1 - \cos \theta } \right) + {1 \over 2}mv_0^2 = {1 \over 2}m{v^2}$.

The radial forces on the right body will give

$T - \cos \theta mg = m{{{v^2}} \over r}$.

Now you need to show that $T > mg$

I don't know if this changes anything, but the initial figure is supposed to look like this
http://www.drcruzan.com/Images/Physics/Atwoods/EqualMassesNoAcceleration.png

 

[Avaron Cooper]

When considering the mass on the right, it does show any vertical displacement. But it shows a horizontal displacement.
Since the string is not considered to be elastic, let's take its length to be a constant.
At the starting point the masses are at equal height so when the mass on the right moves in the horizontal direction, the mass on the left should move up. Again pointing out that the positiin of the mass doesn't change vertically.

 

[haruspex]

When considering the mass on the right, it does show any vertical displacement. But it shows a horizontal displacement.
Since the string is not considered to be elastic, let's take its length to be a constant.
At the starting point the masses are at equal height so when the mass on the right moves in the horizontal direction, the mass on the left should move up. Again pointing out that the positiin of the mass doesn't change vertically.

In the early stages, both masses will accelerate upwards. The question is (at that point), which acceleration is greater?
Consider the FBDs of the two masses when the right hand string is at some small angle theta to the vertical. What are the vertical forces on each?

 

[Avaron Cooper]

I don't understand why the mass on the right side has an acceleration. Is it because theta varies with time and therefore the tension also varies, and the vertical forces are not in equillibrium?

 

[Qwertywerty]

Would the question be valid for any reasonable value of any of the variables ?

 

[haruspex]

I don't understand why the mass on the right side has an acceleration. Is it because theta varies with time and therefore the tension also varies, and the vertical forces are not in equillibrium?

The left hand mass rises, so the tension must exceed mg. The right hand mass experiences the same tension, but, except right at the start, it will not be vertical.

 

[Qwertywerty]

Why does the mass rise ?

 

[Nathanael]

$T - \cos \theta mg = m{{{v^2}} \over r}$

This assumes r is constant, which assumes the other mass isn't moving up or down.
Your conservation of energy equation also assumes the other mass isn't moving up or down (you have no term for the other mass's potential energy).

The equations of motion would be:
$F_r=mg\cos\theta - T =m( \ddot r - r\dot \theta ^2)$
$F_{\theta} = -mg\sin\theta = m(r\ddot \theta +2\dot r \dot \theta)$
We can eliminate T by considering the forces on the other mass: $T-mg=m\ddot r$

These equations can tell you the upwards acceleration of each mass at the initial time, but beyond that they're not very useful unless you know how to solve them (which I certainly don't).If we make the approximation V₀<<r, then cosθ ≈ 1 and sinθ ≈ θ which would turn the equations into:
$2\ddot r \approx r\dot\theta ^2$
$-g\theta \approx r\ddot \theta +2\dot r \dot \theta$
@haruspex or anyone:
Do you think these equations have a simple-ish solution? Or are they not practically solvable?

 

[haruspex]

$2\ddot r \approx r\dot\theta ^2$
$-g\theta \approx r\ddot \theta +2\dot r \dot \theta$
@haruspex or anyone:
Do you think these equations have a simple-ish solution? Or are they not practically solvable?

I don't think the student is expected to solve, or even write down, those equations.
Did you consider the line I suggested in post #13?

 

[Nathanael]

I don't think the student is expected to solve, or even write down, those equations.
Did you consider the line I suggested in post #13?

I don't either, I am personally curious what the solution would look like.
And yes I understood your post #13.

 

[haruspex]

I don't either, I am personally curious what the solution would look like.
And yes I understood your post #13.

Ok. No, the equations look nasty. Note that for the purposes of the question, the quantity of interest is $l-r(1+\cos(\theta))$, where l is the length of the string. I doubt that helps though.