# Laws of motion and circular motion

1. Aug 30, 2011

### a150daysflood

1. The problem statement, all variables and given/known data
[PLAIN]http://img687.imageshack.us/img687/5055/imagelrl.jpg [Broken]

2. Relevant equations

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

3. The attempt at a solution

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

Last edited by a moderator: May 5, 2017
2. Aug 30, 2011

### Hootenanny

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

3. Aug 30, 2011

### 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?

4. Aug 31, 2011

### Hootenanny

Staff Emeritus
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. Sep 4, 2011

### a150daysflood

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

6. Dec 21, 2013

### haha1234

Did I have anything wrong?
mrÏ‰2=Î¼kmgcos30+mgtan60

7. Dec 21, 2013