# Circular motion

1. Sep 6, 2011

### Latios1314

The coefficient of static friction between the slider and the rod is u=0.2. The inclined rod rotates about a vertical axis AB with a constant angular speed, w, as shown in Figure. At the instant shown, the slider is positioned at 0.6 m from B.
(i) What is the acceleration of slider P if it does not slide on the rod?
(ii) Determine the maximum angular speed the rod can have so that the slider P does not slip up the rod. Draw a free body diagram of the slider showing all the forces acting on it in the vertical plane ABC.
(iii) If the angular speed of the rod increases at a rate of 10 rad/s^2 at the instant shown, before the slider starts to slip, what is the tangential contact force exerted by the rod on the slider?

Note: Answer should be: (i) 0.52w^2 (ii) 4.07 rad/s (iii) 5.2 N

The diagram -
http://pdfcast.org/pdf/circular-motion-3 [Broken]

I understand that we must come up with an equation showing the forces acting along BC and another showing the forces acting along the normal of BC.

But can anyone explain to me why the normal force acting on the slider isn't mgsin60?

My tutor has explained the question in class but he didn't elaborate on this part.

Last edited by a moderator: May 5, 2017
2. Sep 6, 2011

### Latios1314

why would there be a resultant force acting in the y-direction?

3. Sep 6, 2011

### Staff: Mentor

That would be the case if there were no acceleration, but here there is a component of acceleration perpendicular to the rod.

4. Sep 6, 2011

### Staff: Mentor

Not sure how you define the y-direction, but the slider is executing circular motion so what direction is it accelerating?

5. Sep 6, 2011

### Latios1314

y direction refers to the normal of BC

Last edited: Sep 6, 2011
6. Sep 6, 2011

### Latios1314

but why is there a acceleration perpendicular to the rod? why is normal force greater than mgsin60?

7. Sep 6, 2011

### Latios1314

At first, I've tried solving the equation by equating the focres perpendicular to BC to be n=mgsin60. but can't seem to understand why it doesnt work out.

8. Sep 6, 2011

### Staff: Mentor

What's the direction of the acceleration?

9. Sep 6, 2011

### Latios1314

direction meaning?

i was referring to why would there be a resultant force normal to BC.

Shouldn't n=mgsin60? why isn't that the case in this scenario?

10. Sep 6, 2011

### Staff: Mentor

Acceleration is a vector. Which way is it pointing? (Hint: Circular motion!)

Yes, and to see why there will be a resultant force component in that direction, one first must understand how things are accelerating.

No, why should it? How did you derive that expression?

11. Sep 6, 2011

### Latios1314

Do you actually mean that the centripetal acceleration acting on the slider is pointed along BC?

Therefore, there are both vertical and horizontal components to the force?

12. Sep 6, 2011

### Latios1314

Does that mean that when we are facing problems regarding circular motion, we must take into account, the force that gave rise to the centripetal force and not the centripetal force itself?

13. Sep 6, 2011

### Staff: Mentor

The centripetal acceleration will have components parallel and perpendicular to BC. (What's the direction of that centripetal acceleration?)
Therefore the force has components parallel and perpendicular to BC.

14. Sep 6, 2011

### Staff: Mentor

I don't understand this question. Rephrase it.

When dealing with any sort of problem using Newton's laws, you need to know the acceleration. In the case of uniform circular motion, the acceleration is centripetal.

15. Sep 6, 2011

### Latios1314

Not really sure what I'm trying to ask too. Was probably very confused.

Can i say that in this case, the normal force acting on the slider and the friction force acting along BC actually gave rise to the resultant centripetal force pointing towards the centre?

16. Sep 6, 2011

### Staff: Mentor

Yes. The sum of all forces acting on the slider give a net force pointing towards the center.

17. Sep 6, 2011

### Latios1314

This is not one of the question. But could you explain to me why the slider would fly out if w is above a certain value?

I understand the the tangential velocity of the sider would increase. But would that lead to?

18. Sep 6, 2011

### Staff: Mentor

There is a maximum value of static friction, which is what holds the slider on. Go too fast, and the friction will not be enough to maintain the required centripetal acceleration.

19. Sep 6, 2011

### Latios1314

would need further help.

tangential acceleration=0.6sin60 x 10

why is this so instead of a=06sin60 x 100?

20. Sep 6, 2011

### Latios1314

Think I've manage to understand it.

Thanks for the great help. My test is next week. Sure hope i'll be able to do well.