Laws of motion and circular motion

  • #1

Homework Statement


[PLAIN]http://img687.imageshack.us/img687/5055/imagelrl.jpg [Broken]


Homework Equations



Static friction = (9.81sin60 x 0.2)N
Centripetal force = mr(w)²



The Attempt at a Solution



I am only abl to get part (i), need hints for part (ii) and (iii).

Thanks in advanced
 
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Answers and Replies

  • #2
Hootenanny
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What can you say about the forcing acting on the slider if it doesn't move, relative to the rod?
 
  • #3
If it doesn't move, I assume the centripetal force= the static friction
Which is (cos30)mrw^2= (0.2)(sin60)9.81

But this doesn't give me the answer given which is 4.07rad/s. Can highlight where did I go wrong?
 
  • #4
Hootenanny
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If it doesn't move, I assume the centripetal force= the static friction
Which is (cos30)mrw^2= (0.2)(sin60)9.81

But this doesn't give me the answer given which is 4.07rad/s. Can highlight where did I go wrong?
You have only considered one component of the centripetal force: the component that acts parallel to the inclined rod. What about the other component that acts perpendicular to the inclined rod?
 
  • #5
Hi, how do I go abt finding the force by rod on the slider in part(iii)?
 
  • #6
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You have only considered one component of the centripetal force: the component that acts parallel to the inclined rod. What about the other component that acts perpendicular to the inclined rod?

Did I have anything wrong?
mrω2=μkmgcos30+mgtan60
 
  • #7
adjacent
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Did I have anything wrong?
mrω2=μkmgcos30+mgtan60
@haha1234, this thread is about 2 years old and @Hootenanny is no longer active on PF.
 

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