Physics of Indiana Jones rope swing

AI Thread Summary
The discussion revolves around solving a physics problem related to Indiana Jones swinging from a rope, modeled as a simple pendulum. The key equations mentioned include the period of the pendulum, T = 2π√(L/g), and the angle calculation, θ = θ₀ cos((2π/T) t). Participants express confusion about applying these equations correctly, particularly in calculating the angle after 1.290 seconds. The period calculation appears to be accurate, but there is a misunderstanding in the application of the angle equation. Clarification is needed on how to correctly use the initial angle and time to find the new angle after the specified duration.
melissa_y
Messages
17
Reaction score
0
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!
 
Physics news on Phys.org
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.
 
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!
 
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)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top