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Gravity of Moon & Roller Coaster Proof Help

  • Thread starter Klinger
  • Start date
  • #1
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I would appreciate any hints or help in solving the following two physics problems. At this point I am stuck after trying to solve the problems for a while.

1. Apollo astronauts hit a golf ball on the moon 180 meters and 30 meters on earth. Assume that the swing, launch angle, etc are the same on the moon and earth. Assume no air resistance.

We know that Vxo and Vyo is the same on both the moon and earth (Vxo is the initial velocity in the x direction). We also know that tmoon = tearth *(180/30).

I try substituting these values into the kinematic equations for projectile motion but get stuck. Vy are different for earth and moon and thus there are too many unknowns.

2. Show that on a roller coaster with a circular vertical loop the difference in your apparent weight at the top of the loop and the bottom of the loop is 6g's. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn't depend on the size of the loop or how fast you go through it. Assume that roller coaster starts traveling from height h and the radius of the loop is R.

My approach was to calculate the ratio of the centripidal acceleration as it exits the loop divided by the centripidal acceleration at the top of loop. I calculate the decease in velocity due to the elevation gain of 2R. But I don't seem to get the expected answer.

Thanks for any help.

:bugeye:
 

Answers and Replies

  • #2
698
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for your first question, are you asked for the gravity on the moon?
then you can use this:
On earth:
[tex] V_{earth}^{2} = V_{o\ earth}^{2} + 2ad [/tex]
[tex] V_{earth}^{2} - V_{o\ earth}^{2} = 2a(1) [/tex]
[tex] V_{earth}^{2} = V_{o\ earth}^{2} = 2a [/tex]

On moon:
[tex] V_{moon}^{2} = V_{o\ moon}^{2} + 2ad [/tex]
[tex] V_{moon}^{2} - V_{o\ moon}^{2} = 2a(\frac{30m}{180m}) [/tex]
[tex] V_{moon}^{2} = V_{o\ moon}^{2} = \frac{1}{3}a [/tex]

Since the initial and final speed are the same, we can eqate the two:
[tex]2a = \frac{1}{3}a [/tex]
[tex]a_{moon} = \frac{1}{6}a_{earth} [/tex]
[tex]g_{moon} = \frac{1}{6}g_{earth} [/tex]

I think you can do the second one yourself.
 
  • #3
4
0
A Couple of Clarification Questions

I have a couple of questions:
-- in the 2nd "earth" equation why did you set d=1?
-- in the 2nd "moon" equation why did you set d=30/180?

Nenad said:
for your first question, are you asked for the gravity on the moon?
then you can use this:
On earth:
[tex] V_{earth}^{2} = V_{o\ earth}^{2} + 2ad [/tex]
[tex] V_{earth}^{2} - V_{o\ earth}^{2} = 2a(1) [/tex]
[tex] V_{earth}^{2} = V_{o\ earth}^{2} = 2a [/tex]

On moon:
[tex] V_{moon}^{2} = V_{o\ moon}^{2} + 2ad [/tex]
[tex] V_{moon}^{2} - V_{o\ moon}^{2} = 2a(\frac{30m}{180m}) [/tex]
[tex] V_{moon}^{2} = V_{o\ moon}^{2} = \frac{1}{3}a [/tex]

Since the initial and final speed are the same, we can eqate the two:
[tex]2a = \frac{1}{3}a [/tex]
[tex]a_{moon} = \frac{1}{6}a_{earth} [/tex]
[tex]g_{moon} = \frac{1}{6}g_{earth} [/tex]

I think you can do the second one yourself.
 

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