I think my answer should be right, but it isn't energy problem

  • Thread starter holezch
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Homework Statement



A light rigid rod of length L has a mass m attached to its end, forming a simple pendulum. It is inverted and then released (I don't know what that's supposed to mean). What are:

a) the speed v at its lowest point
b) the tension T in the suspension at that instant

Homework Equations





The Attempt at a Solution



at the beginning, when it is dropped down, we should have K = 0 and U = mgL

at the bottom when it is finished dropping, we should have K = 1/2mv^2 U = 0

1/2 mv^2 = mgl

v = sqrt(2gL)

but the book says 2sqrt(gl)...

what did I do wrong?

thanks
 

Answers and Replies

  • #2
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I'm guessing "inverted" means the mass is a height L above the pivot point, so it's height above its lowest point is actually 2L rather than L
 
  • #3
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WOW. Thanks so much, I would never have known...

What about a "suspension"?

b) the tension T in the suspension at that instant..

Shouldn't it just be the negative weight? what answer book says 5mg.. :S

Thanks
 
  • #4
PhanthomJay
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WOW. Thanks so much, I would never have known...

What about a "suspension"?

b) the tension T in the suspension at that instant..

Shouldn't it just be the negative weight? what answer book says 5mg.. :S

Thanks
the 'tension T in the suspension at that instant' means 'what is the tension in the rod when the mass is suspended by the rod at the low point of its motion (that is, upon returning to its original position before it was inverted, but this time with a speed associated with it). This becomes a centripetal acceleration question, the tension is not the negative of the weight, because it is accelerating inwardly. The sum of forces in the y direction is not 0.
 

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