# Indiana Jones rope swing physics

• zhenyazh
In summary, The question revolves around Indiana Jones swinging on a rope, treated as a simple pendulum, with a distance of 31.0 m between the pivot point and his center of mass. After starting from rest at an angle θ = 19.0o, the question asks for the value of θ after 1.27 s. The formula used is x=Acos(wt+fi), where A=19 deg, t=1.27 s and w=sqrt(g/l), with l being 31. There is a discussion about using degrees instead of meters, and using fi=0 for the calculation.
zhenyazh
Hi,
I am preparing for a test and have the following question.
there is the image attached too.

Indiana Jones is swinging from a rope. The distance between the pivot point and his center of mass is 31.0 m. He begins swinging from rest at an angle θ = 19.0o as shown in the figure. Assuming that Indiana and the rope can be treated as a simple pendulum, what is the value of θ after 1.27 s (in degrees)?

so i this is a harmonic motion question. as far as i understand i can use the
formula of location (x=...) but use angles to represent the location and thus get the answer. but somehow i get the wrong answer all the time.
could some one show me how it is done so i can see where i am wrong?

thanks

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• indiana.JPG
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Hi zhenyazh!
zhenyazh said:
… so i this is a harmonic motion question. as far as i understand i can use the
formula of location (x=...) but use angles to represent the location and thus get the answer. but somehow i get the wrong answer all the time.
could some one show me how it is done so i can see where i am wrong?

thanks

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

ok.
so indiana starts from the farthest point and i want to work in degrees and not meters t to find position.
so i say x=Acos(wt+fi)
where A=19 deg, t=.127 and w=sqrt(g/l) where l is 31 if i understand the question correctly.
the answer i get is wrong.

Hi zhenyazh!

(have an omega: ω and a phi: φ and a square-root: √ )
zhenyazh said:
ok.
so indiana starts from the farthest point and i want to work in degrees and not meters t to find position.
so i say x=Acos(wt+fi)
where A=19 deg, t=.127 and w=sqrt(g/l) where l is 31 if i understand the question correctly.
the answer i get is wrong.

Isn't t = 1.27 ? And what did you use for φ ?

about the time yes you are right, it is a type.
about fi, i used zero since i use cos for my calculation and start from the right amplitude

## 1. How does Indiana Jones swing on a rope without losing momentum?

According to the laws of physics, momentum is conserved unless an external force acts on an object. In Indiana Jones' case, he maintains his momentum by using the force of his legs and body to push off the swing at the bottom of the arc, just like a pendulum.

## 2. Can Indiana Jones really swing on a rope like that in real life?

In reality, swinging on a rope like Indiana Jones would require an immense amount of upper body strength and perfect timing. It is possible for highly trained individuals, but it is not a practical or efficient method of transportation.

## 3. How does the length of the rope affect Indiana Jones' swing?

The length of the rope affects the period and speed of the swing. A longer rope will result in a longer period and slower swing, while a shorter rope will result in a shorter period and faster swing. Indiana Jones would need to adjust his timing and technique accordingly.

## 4. Can Indiana Jones change the direction of his swing while on the rope?

Yes, Indiana Jones can change the direction of his swing by altering the angle at which he jumps off the rope at the bottom of the arc. By pushing off at an angle, he can change the direction of his momentum and swing in a different direction.

## 5. How does friction play a role in Indiana Jones' rope swing?

Friction is an important factor in Indiana Jones' rope swing. If there is too much friction, he may not be able to swing as high or as far. However, if there is not enough friction, he may slip off the rope and fall. The material and condition of the rope and the surface he is swinging over can affect the amount of friction present.

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