|
Forums:
Climbing Information:
The Lab:
Re: [ptlong2] Pendulum fall speed:
Edit Log
|
|
jt512
Sep 1, 2010, 1:25 AM
Views: 20813
Registered: Apr 12, 2001
Posts: 21904
|
ptlong2 wrote: patto wrote: For a perfect pendulum with a dynamic rope the speed is less. The tension on the rope will induce stretch the rope and absorb energy. Yes, less. But the stretch also lengthens the fall which adds energy to the system, nearly enough to make it a wash. For a Hookian rope the difference is maybe a percent or two, depending on the rope modulus you select. If that's true then can you solve for the peak tension as follows? The potential energy (PE) of the fall is converted to kinetic energy (KE) and strain energy (SE) in the rope. That is, (1) PE = KE + SE . (2) PE = mgh = mg(L + s) , where L is the length of the unstretched rope and s is the maximum rope stretch in the fall. (3) KE = (1/2)mv² , (4) SE = (k/2L)s² (from rgold's paper), where k is the rope modulus. Substitute (2), (3), and (4) into (1), to give (5) mg(L + s) = (1/2)mv² + (k/2L)s² . From rgold's paper (6) s = TL/k , where T is the maximum rope tension. Now, substitute (6) into (5), compute v assuming an ideal pendulum, and solve for T. Jay
(This post was edited by jt512 on Sep 1, 2010, 1:26 AM)
|
|
Edit Log:
|
Post edited by jt512
() on Sep 1, 2010, 1:26 AM
|
|
|
|
|
|