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Physics of Indiana Jones rope swing

  • Thread starter melissa_y
  • Start date
  • #1
17
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Indiana Jones!!!!!

My physics class had this problem for our homework set this week and nobody has figured it out. We asked for help from our TAs today and they didn't know how to help us either. If someone could look at this problem and tell me what they think I would really appreciate it.

Indiana Jones is swinging from a rope. The distance between the pivot point and his center of mass is 31.00m. He begins swinging from rest at an angle theta=18.00. Assuming that Indiana and the rope can be treated as a simple pendulum, what is the value of theta after 1.290s (in degrees)?

Ok so this is the farthest that any of us have gotten so far...

T= 2pi square root of L/g
theta= theta,o cos((2pi/T) t)

I think these are the equations we are supposed to use but I am still really confused on how to solve this problem. If someone could direct me in the right way and give me the right idea my WHOLE physics class would really appreciate it. Thanks!
 

Answers and Replies

  • #2
2,209
1
Looks like you are applying SHM principles which is good. Find the period and angular velocity and from there just plug it into that bottom equation.
 
  • #3
17
0
Ok so this is what I've done and I am not gettingthe problem right still....

I did T = 2pi (square root of L/g)

I took that T/2 = 5.5875

Then i set up this equation

5.5875/1.290 = 18/x

Then I solve for x and subtract the x from the theta given. I'm not sure what i"m doing wrong!
 
  • #4
OlderDan
Science Advisor
Homework Helper
3,021
2
melissa_y said:
Ok so this is what I've done and I am not gettingthe problem right still....

I did T = 2pi (square root of L/g)

I took that T/2 = 5.5875

Then i set up this equation

5.5875/1.290 = 18/x

Then I solve for x and subtract the x from the theta given. I'm not sure what i"m doing wrong!
The period calculation looks correct. Your last equation is not correct. Why are you not using the equation for theta from your original post?

theta= theta,o cos((2pi/T) t)
 

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