Force in circular motion

  • #1
Just wondering if anyone can check over my answer for a question.

Q) A 1.7kg object is swung on the end of a 0.6m string in a vertical circle. The object does one revolution every 1.1s. What is the tension in the string at the top of the circle? and the bottom?

A) m=1.79kg r=0.6m t=1.1s

[tex]\omega[/tex] = 2[tex]\pi[/tex]/1.1 = 5.711986643ms-1

F=mr[tex]\omega[/tex]2 = 1.7*0.6*5.7119866432=33.2793N
minus force due to gravity 1.7*9.8=16.6193N
= 16.71N at top

33.2793N + 16.6193N = 50N at the bottom

Many thanks

p.s( the omega is not to the power...its just the way it came out after i wrote it)
 

Answers and Replies

  • #2
gneill
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What keeps the angular velocity constant? Shouldn't the mass' kinetic energy decrease as it climbs higher in the gravitational field (trading KE for PE)? Or are we to assume that whatever is swinging the string around is compensating for this?
 
  • #3
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The way I see it, there's nothing wrong. I got the same answers as you.
 
  • #4
Doc Al
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The way I see it, there's nothing wrong.
If you assume that ω is constant, as gneill points out.
 
  • #5
gneill
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The way I see it, there's nothing wrong. I got the same answers as you.
Compare the KE for the assumed constant-speed motion to the energy gained or lost due to change in PE as an object rises or falls through the same vertical distance (top and bottom of the loop). How do the magnitudes compare? Is is safe to assume constant speed?
 

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