Speed and apparent weight of person in reverse bungee jumping

[songoku]

Homework Statement

This is set up of reverse bungee jumping (see picture below). The total mass is 500 kg, the original length of cord is 25 m and total elastic potential energy at ground level is 0.6 MJ. Find:
a. speed of capsule when the cords first become loose
b. position where the apparent weight of the person inside it will be the greatest

Relevant Equations

Conservation of energy

Newton's Law

a. To find the speed, I need to find the height where the cords first become loose, which is when the cord is 25 m long.
$$h=30-\sqrt{25^2 - 5^2}$$

But the teacher's working is $h=30-\sqrt{25^2 - 4^2}$

How to get 4?

b. My idea is to use Newton's 2nd law so I draw free body diagram of the person inside the capsule. There are two forces, normal force acting upwards and weight acting downwards. The apparent weight is the normal force acting on the person. The equation of motion is:
$$N-W = m.a$$
$$N=W+m.a$$

So the apparent weight is the largest when the acceleration is the largest. How to know the position where the acceleration is the largest?

Thanks

 

[Filip Larsen]

How to get 4?

Perhaps there is some information in the original problem description on the width of the capsule, or more specifically, the distance between the two cord attachment points on the capsule?

 

[haruspex]

the original length of cord is 25 m

Should that read "length of each cord"?

How to know the position where the acceleration is the largest?

A good first step would be to write an equation for the upward motion. Either start with forces and accelerations or look for a conservation principle that might help.

 

[songoku]

Perhaps there is some information in the original problem description on the width of the capsule, or more specifically, the distance between the two cord attachment points on the capsule?

You are correct. I just asked my teacher and he said there should be information about effective diameter of the capsule, which is 2 m. Now I understand how to get 4 (I draw the diagram when the length of each cord is 25 m)

Should that read "length of each cord"?

Yes, my bad. I am sorry

A good first step would be to write an equation for the upward motion. Either start with forces and accelerations or look for a conservation principle that might help.

Would the equation for upward motion be the one I wrote in #1?
$$N=W+ma$$

For conservation principle, the one I think is conservation of energy. I think I need to find the position where the KE is the greatest since it corresponds to highest speed and maybe highest acceleration (I am not so sure).

At start of motion, KE is zero then it will increase until the cord first becomes loose since there is no more elastic potential energy can be converted into KE so the largest apparent weight will be at position of $30 - \sqrt{25^2 - 4^2}$ above the ground.

Is this correct? Thanks

 

[haruspex]

highest speed and maybe highest acceleration

No, highest speed would be zero acceleration. If acceleration is positive then a higher speed is coming, and if it is negative then a higher speed was achieved in the past.

KE is zero then it will increase until the cord first becomes loose

That is not true either. By the time the cord is loose the speed will have dropped.

Yes, use conservation of work, but write the general expression for the total work at any given height.

 

[songoku]

Yes, use conservation of work, but write the general expression for the total work at any given height.

Total work = $\Delta GPE+\Delta KE-\Delta EPE$ (EPE = elastic potential energy)

$$=mg(30-\sqrt{L^2-4^2})+\frac{1}{2}mv^2+2 \times \frac{1}{2}k(L-25)^2$$

Is that correct? If yes, what should I do next to know about apparent weight?

Thanks

 

[haruspex]

Total work = $\Delta GPE+\Delta KE-\Delta EPE$ (EPE = elastic potential energy)

$$=mg(30-\sqrt{L^2-4^2})+\frac{1}{2}mv^2+2 \times \frac{1}{2}k(L-25)^2$$

Is that correct? If yes, what should I do next to know about apparent weight?

Thanks

Sorry, but having thought about it a bit more I see a much easier way.
As you noted in post #4, max apparent weight corresponds to max normal force and thus to max acceleration. What vertical forces act on the person+capsule? Which of these is variable? At what position is it maximised?

 

[songoku]

Sorry, but having thought about it a bit more I see a much easier way.
As you noted in post #4, max apparent weight corresponds to max normal force and thus to max acceleration. What vertical forces act on the person+capsule? Which of these is variable? At what position is it maximised?

Vertical forces are two tensions directed upwards and weight. The equation will be:
$$2T \cos \theta -W=m.a$$
where $\theta$ is angle between the cord and vertical

To get maximum $a$, it means that both $T$ and $\cos \theta$ should be maximum. $T$ is maximum when the cord has maximum extension and $\cos \theta$ is maximum when the angle is the smallest so the position where the apparent weight is the largest is at the lowest position when the capsule starts to move.

Is that correct? Thanks

 

[haruspex]

Vertical forces are two tensions directed upwards and weight. The equation will be:
$$2T \cos \theta -W=m.a$$
where $\theta$ is angle between the cord and vertical

To get maximum $a$, it means that both $T$ and $\cos \theta$ should be maximum. $T$ is maximum when the cord has maximum extension and $\cos \theta$ is maximum when the angle is the smallest so the position where the apparent weight is the largest is at the lowest position when the capsule starts to move.

Is that correct? Thanks

Yes.

 

[songoku]

Thank you very much Filip Larsen and haruspex