Velocity of Pendulum at a certain point

AI Thread Summary
Tarzan swings from a 23.0 m vine, descending 3.2 m, and the problem focuses on calculating the maximum tension in the vine at the bottom of the swing. The key equations involve force summation and radial acceleration, with the maximum tension occurring when Tarzan is at the lowest point. The discussion emphasizes using conservation of energy to find the velocity at that point, as potential energy converts to kinetic energy during the descent. The initial kinetic energy is zero since Tarzan starts from rest, and the reasoning aligns with energy conservation principles. The participant successfully solved the problem and plans to return for further assistance on similar topics.
grantrudd
Messages
7
Reaction score
0

Homework Statement



Tarzan, who weighs 668 N, swings from a cliff at the end of a convenient vine that is 23.0 m long (see the figure). From the top of the cliff to the bottom of the swing, he descends by 3.2 m.
If the vine doesn't break, what is the maximum magnitude of the tension in the vine?

http://capa-new.colorado.edu/hrw-lib/hrwpictures/8-38.jpg

Homework Equations



F=ma

a_radial=v^2/r

The Attempt at a Solution



i know that the max tension is at the bottom, so i made a force diagram at the bottom of the path, and summed the forces to get

F=T-mg
mv^2/r + mg= T

i found the mass, but my only missing variable is the velocity of the "pendulum."

i am completely stuck on this part. Thanks for any help!
 
Physics news on Phys.org
Welcome to PF!

Can you think of anyways you can get V from this information "From the top of the cliff to the bottom of the swing, he descends by 3.2 m."
 
As energy is not lost nor created, Tarzan speed (kynetic energy) must come from another form of energy. In this case, he has descended 3.2m...
 
thanks for the welcome!

i have been thinking about it on the next homework problem actually, and i think it has something to do with conservation of momentum. something like:

1/2(mv^2)+mgy= 1/2(mv^2) +No potential at the bottom.

the initial velocity is 0, so the first initial kinetic energy is 0. is this reasoning right? or does he have some kinetic at the top?
 
i guess that was right. i got the problem right, so that should help me on a few other problems. i will probably be back here asking more questions later though!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top