1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Circular motion

  1. Sep 6, 2011 #1
    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. jcsd
  3. Sep 6, 2011 #2
    why would there be a resultant force acting in the y-direction?
     
  4. Sep 6, 2011 #3

    Doc Al

    User Avatar

    Staff: Mentor

    That would be the case if there were no acceleration, but here there is a component of acceleration perpendicular to the rod.
     
  5. Sep 6, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Not sure how you define the y-direction, but the slider is executing circular motion so what direction is it accelerating?
     
  6. Sep 6, 2011 #5
    y direction refers to the normal of BC
     
    Last edited: Sep 6, 2011
  7. Sep 6, 2011 #6
    but why is there a acceleration perpendicular to the rod? why is normal force greater than mgsin60?
     
  8. Sep 6, 2011 #7
    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.
     
  9. Sep 6, 2011 #8

    Doc Al

    User Avatar

    Staff: Mentor

    What's the direction of the acceleration?
     
  10. Sep 6, 2011 #9
    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?
     
  11. Sep 6, 2011 #10

    Doc Al

    User Avatar

    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?
     
  12. Sep 6, 2011 #11
    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?
     
  13. Sep 6, 2011 #12
    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?
     
  14. Sep 6, 2011 #13

    Doc Al

    User Avatar

    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.
     
  15. Sep 6, 2011 #14

    Doc Al

    User Avatar

    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.
     
  16. Sep 6, 2011 #15
    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?
     
  17. Sep 6, 2011 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. The sum of all forces acting on the slider give a net force pointing towards the center.
     
  18. Sep 6, 2011 #17
    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?
     
  19. Sep 6, 2011 #18

    Doc Al

    User Avatar

    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.
     
  20. Sep 6, 2011 #19
    would need further help.

    tangential acceleration=0.6sin60 x 10

    why is this so instead of a=06sin60 x 100?
     
  21. Sep 6, 2011 #20
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Circular motion
Loading...