Elastic pendulum energy problem

[holezch]

Homework Statement

Suppose that the string in fig 8-12 (just a pendulum released from the side and falls to the bottom) is very elastic, made of rubber, say, and that the string is unextended at length l when the ball is released.

a) explain why you would expect the ball to reach a low point greater than a distance l below the point of suspension.

b) show, using dynamic and energy considerations, that if \Deltal is small compared to l, the string will stretch by an amount \Deltal = 3 mg/k, where k is the assumed force constant of the string.

Homework Equations

The Attempt at a Solution

for a, I suppose that the weight would displace the mass further down before the force of the string -kx "pulls" it back. but I'm not entirely sure..

for b) I tried to work out the change in K = Work from weight + work from string(spring like). I think if the change in l is small compared to l, then I can just say change in l + l = l, approximately. besides that, I'm not really sure.. any help please? thanks

 

[holezch]

I stopped thinking about this but now came back to it.. please, can someone help? My argument is that the weight will pull it down until the force done by the rubber band (-kx) is >= than the weight.

For part b, I got pretty close:


if \DeltaL is relatively small to L, then L+\DeltaL is approximately L.

So, just as it is released, the system has an energy level of mgL + \DeltaL, approximately mgL. Then the system at its lowest point will have K approx = mgL. At the point just as the rubber band is about to stretch, mgL = K + mg\DeltaL. Then, K = mgL - mg\DeltaL.

So then we have

\Deltak = \Sigma Wc


or approximately,

mgL - mgL + mg\DeltaL = -k\DeltaL^2 - mg\DeltaL

2mg\DeltaL = -k\DeltaL^2

then we have 2mg = -K\DeltaL

and \DeltaL = 2mg/-K

but the answer is 3 mg / K.. What went wrong?

thanks

 

[holezch]

okay, I realized my error - there is no "At the point just as the rubber band is about to stretch, mgL = K + mgL. Then, K = mgL - mgL." the rubber band stretches delta L over the fall.
back to the drawing board..

 

[holezch]

can anybody help? please?

 

[holezch]

still looking for help..

 

[ideasrule]

for a, I suppose that the weight would displace the mass further down before the force of the string -kx "pulls" it back. but I'm not entirely sure..

Write out Newton's second law for the radial (towards the center) direction. You'll see that the tension in the string has to balance out the radial component of gravity and provide the centripetal acceleration. What happens to both as the bob moves downwards?

for b) I tried to work out the change in K = Work from weight + work from string(spring like). I think if the change in l is small compared to l, then I can just say change in l + l = l, approximately. besides that, I'm not really sure.. any help please? thanks

Try writing out the equation for K = Work from weight + work from string in terms of l, delta-l, k, m, etc.

 

[ideasrule]

I don't quite understand the calculations you did for b, so here's a hint: go back to the equation you wrote in part a) for the radial direction. You'll see that if you know the speed of the pendulum at the bottom, you can solve for tension. From there you can solve for delta-L. How do you use the conservation of energy to solve for speed?

 

[holezch]

I don't quite understand the calculations you did for b, so here's a hint: go back to the equation you wrote in part a) for the radial direction. You'll see that if you know the speed of the pendulum at the bottom, you can solve for tension. From there you can solve for delta-L. How do you use the conservation of energy to solve for speed?

thanks a lot - I appreciate it. but don't we already know the kinetic energy at the bottom? and what's the idea behind "that if that if LaTeX Code: \\Delta l is small compared to l".. should we consider L + delta L to just be L? I tried to solve for delta-L by sum of work = change in K.. where we know the bottom K

thanks again! :)

 

[ideasrule]

If you write out the equation I mentioned, you'd see that T=mg+mv^2/L at the bottom. delta-L=T/k, so you need to find v in order to find delta-L. Yes, you know the kinetic energy at the bottom, so finding v should be easy.

Also, yes, you can consider L+delta-L to equal L when computing v.

 

[holezch]

thanks for replying. when I calculate the work = change in K, I need to calculate the work done by the weight, the spring force and the tension right? is that why I need to know about the tension? (other than the first part of the question)