Conservation of Energy in a System of Three Balls and Two Pulleys

[Nexus99]

Homework Statement

Two balls of mass m are bounded by two wires of equal lenght to a third ball of mass 2m.
The wire passes on two pulleys, that are distant 2l. The ball of mass 2m, initially firm at the same
height of the pulleys and at the same distance from them, is dropped until it reaches the floor, distant h from the initial height. Which is its velocity in that moment?

Relevant Equations

Conservation of energy

i tried with conservation of energy.

$E_i = 0$
$E_f = 2 m v^2 - 2mgh - 2mg(\sqrt{h^2 + l^2} - l)$
$v = \sqrt{g(h - l + \sqrt{h^2 + l^2})}$
Is it right?

 

[Chestermiller]

The potential energy of the two masses increases, not decreases. And you left out the final kinetic energy of the two masses.

 

[Nexus99]

The potential energy of the two masses increases, not decreases. And you left out the final kinetic energy of the two masses.

Ok so P.E of the two massese is $2mg(\sqrt{h^2 + l^2} - l)$
I think that kinetic energy is correct:
$\frac{1}{2} 2m v^2 + 2 \frac{1}{2} mv^2 = 2mv^2$ is it ok?

 

[TSny]

Ok so P.E of the two massese is $2mg(\sqrt{h^2 + l^2} - l)$
I think that kinetic energy is correct:
$\frac{1}{2} 2m v^2 + 2 \frac{1}{2} mv^2 = 2mv^2$ is it ok?

You've assumed that all three masses have the same speed when the middle mass reaches the floor.

 

[Nexus99]

You've assumed that all three masses have the same speed when the middle mass reaches the floor.

I thought so, which is the relationship between the speeds?

 

[TSny]

I thought so, which is the relationship between the speeds?

The middle mass has a velocity that is vertically downward. This velocity has components parallel and perpendicular to the portion of the string on the left that runs from the pulley to the middle mass.

 

[Nexus99]

So, $v_{//_{2m}} = - v_m$ ?
I tried also a new approach
$v_{2m} = \dot{h}$
$v_{m} = \frac{d}{dt} (\sqrt{h^2 + l^2} - l) = \dot{h} \frac{h}{\sqrt{h^2 + l^2}}$
$0 = m(\dot{h} \frac{h}{\sqrt{h^2 + l^2}})^2 + m \dot{h}^2 - 2mgh + 2mg(\sqrt{h^2 + l^2} - l)$
$\dot{h} = \sqrt{[\frac{2g(h+l - \sqrt{h^2 + l^2}}{2h^2 + l^2}](h^2 + l^2)}$

Is it right?

 

[Chestermiller]

Ok so P.E of the two massese is $2mg(\sqrt{h^2 + l^2} - l)$
I think that kinetic energy is correct:
$\frac{1}{2} 2m v^2 + 2 \frac{1}{2} mv^2 = 2mv^2$ is it ok?

No. The velocity of the two m masses is not equal to the velocity of the 2m mass. There is a geometric factor.

 

[TSny]

So, $v_{//_{2m}} = - v_m$ ?
I tried also a new approach
$v_{2m} = \dot{h}$
$v_{m} = \frac{d}{dh} (\sqrt{h^2 + l^2} - h) = \dot{h} \frac{h}{\sqrt{h^2 + l^2} - h}$
$0 = m(\dot{h} \frac{h}{\sqrt{h^2 + l^2} - h})^2 + m \dot{h}^2 - 2mgh + 2mg(\sqrt{h^2 + l^2} - l)$
$\dot{h} = \sqrt{[\frac{2g(h+l - \sqrt{h^2 + l^2}}{2h^2 + l^2}](h^2 + l^2)}$

Is it right?

All looks good to me.

 

[Chestermiller]

So, $v_{//_{2m}} = - v_m$ ?
I tried also a new approach
$v_{2m} = \dot{h}$
$v_{m} = \frac{d}{dh} (\sqrt{h^2 + l^2} - h) = \dot{h} \frac{h}{\sqrt{h^2 + l^2} - h}$
$0 = m(\dot{h} \frac{h}{\sqrt{h^2 + l^2} - h})^2 + m \dot{h}^2 - 2mgh + 2mg(\sqrt{h^2 + l^2} - l)$
$\dot{h} = \sqrt{[\frac{2g(h+l - \sqrt{h^2 + l^2}}{2h^2 + l^2}](h^2 + l^2)}$

Is it right?

No. If $L=\sqrt{h^2+l^2}$, the velocity of the 1m masses is dL/dt.

 

[Nexus99]

No. If $L=\sqrt{h^2+l^2}$, the velocity of the 1m masses is dL/dt.

Yes, sorry i derived dt but i wrote dh

 

[TSny]

No. If $L=\sqrt{h^2+l^2}$, the velocity of the 1m masses is dL/dt.

Yes, I didn't catch that. Thanks. Looks like the final answer is correct.