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Shooting distance

  • Thread starter robax25
  • Start date
  • #1
182
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Homework Statement


A catapult has got a thin slender cylindrical lever of 4 m length and 50 kg mass (ignore the
detailed spoon kind of shape). Loading the spring mechanism took 2 minutes with a constant
power of 2000 W. A rock of 150 kg is put at the tip of the lever in the horizontal position. In
order to shoot the rock, the lever is rotating from the horizontal to the vertical position. The
rock is gaining potential as well as motion energy. Calculate the maximum shooting distance!

Homework Equations



mgh=0.5*m*v²
x=vt

The Attempt at a Solution


x=76.79m[/B]
 

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Answers and Replies

  • #2
182
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question is that either I am right or wrong? I do not have any solution.
 
  • #3
122
35
Your methodology looks OK but 1) the moment of inertia of a rod about one end is mL^2 / 3 and the moment of inertia of the rock is ML^2; 2) the potential energy of the lever and rock in the vertical position is mgL/2 + MgL. Make those changes and recalculate the distance and I'll let you know if we agree.
 
  • #4
haruspex
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Calculate the maximum shooting distance!
the lever is rotating from the horizontal to the vertical position.
It is not entirely clear that the rock is released when the arm is vertical. A greater range may be achieved by releasing it earlier, when the rock's velocity has an upward component. Indeed, if there is nothing to prevent it, it might naturally launch sooner. However, we would not then know how much unused energy remains in the spring.
 
  • #5
Ray Vickson
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question is that either I am right or wrong? I do not have any solution.
PF standards are that you avoid posting images, but type out your work (reserving images for diagrams, etc). Consult the Guidelines for more on this issue.
 

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