Laws of motion and circular motion

[a150daysflood]

Homework Statement

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

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

 

[Hootenanny]

What can you say about the forcing acting on the slider if it doesn't move, relative to the rod?

 

[a150daysflood]

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?

 

[Hootenanny]

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?

 

[a150daysflood]

Hi, how do I go abt finding the force by rod on the slider in part(iii)?

 

[haha1234]

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=μ_kmgcos30+mgtan60

 

[adjacent]

Did I have anything wrong?
mrω²=μ_kmgcos30+mgtan60

@haha1234, this thread is about 2 years old and @Hootenanny is no longer active on PF.