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!

Centrifugal and centripetal forces on a half pipe

  1. Nov 11, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem is as follows:
    upload_2015-11-11_11-17-21.png
    A skater started at the top of a halfpipe that makes a 40 degree angle with the horizontal. He is now about halfway down the pipe and represented by the rectangle. His center of gravity is represented by Z, the midpoint of the circular movement he's completing in the halfpipe is represented by M. The radius of said circle is 2,8 meters, his center of gravity is 0.7 meters off the surface of the ramp. The question is to calculate the normal force exerted by the ramp on the skater. The skater weighs 61 kg and his radial speed at the moment represented in the figure is 3.2 rad/s
    2. Relevant equations

    Fz = m*g*cos α
    Fmpz = (m*v2)/r = m*ω2*r

    3. The attempt at a solution

    The centripetal force would be the one allowing the skater to complete a circular motion, the gravity is pulling the skater straight down but has components that aid the skater in his downward motion (not important here) and a component that is perpendicular to the surface of the ramp. In my opinion the neutral force would then need to be added to the centripetal force so both counteract the perpendicular component of the gravity in essence:

    Fn + Fmpz = Fz-component leads to: Fn = Fz-component - Fmpz
    and the radius is the total radius of the circle minus the elevation of the center of gravity of the skater = 2.8-0.7 = 2.1 m

    Fz = 61 * 9.81 (gravitational acceleration) * cos 40 = 484 N
    Fmpz = 61 * 3.22*2.1 = 1311 N

    However this results in a negative neutral force: Fn = 484-1311 = -827

    The answer provided however states that the neutral force is equal to the perpendicular gravity component added to the centripetal force (Fn = Fmpz + Fzcomponent) but that would mean the centripetal force is actually pointed away from the center of the circle (more like a centrifugal force?), I've always been taught not to work with the centrifugal force as it is not a 'real force'? Is that wrong?
     
  2. jcsd
  3. Nov 11, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I assume where you refer to 'neutral' force you mean normal force.
    Centripetal force is not an applied force. It is the radial component of the resultant of all the applied forces. It is better to think in terms of centripetal acceleration: ##\Sigma F_{radial}=m a_{centripetal}##.

    It is ok to work with centrifugal force instead so long as you understand it is a 'fictitious' force. That is, it is a force perceived in an accelerating reference frame to account for the apparent violation of Newton's laws. It is, however, an applied force, so is include in the sum of forces.
    If you have been taught to spurn the use of non-inertial frames, let's stick with that.
     
  4. Nov 11, 2015 #3
    Yes, sorry bit of translation issue I do mean the normal force. As for your statement, the centripetal force in this problem should then be composed of other forces (or components of other forces), but it should still be directed at the midpoint of the circle, no? So given that the normal force is directed upwards (perpendicular) from the ramp, the centripetal force and normal force should be going in the same direction, rather than opposite directions as the (correct) answer to the problem would suggest?
     
  5. Nov 11, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, the normal force and resultant centripetal force both point towards the centre of rotation here. Are you saying the given answer contradicts that? I don't see how.
     
  6. Nov 11, 2015 #5
    The correct answer to the problem states that: Fn = Fmpz + Fzcomponent, this answer does appear to be correct given that my own solution results in a negative normal force. Given the correct answer I took it to mean that the centripetal force and the Fz component are going in the same direction, and that therefore the centripetal force and normal force are going in opposite directions. Which made me think there not actually calculating a centripetal but rather a centrifugal force?
     
  7. Nov 11, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    As I wrote, it's clearer to write it as resultant = sum of applied. This gives Fc=FN+mg cos(40), where up is positive and g takes a negative value. If you prefer to work with g taking a positive value then it's Fc=FN-mg cos(40), or FN=Fc+mg cos(40). In your notation, Fn = Fmpz + Fzcomponent.
     
  8. Nov 11, 2015 #7
    I tried to draw everything relevant into the original figure:
    upload_2015-11-11_12-49-14.png
    The centrifugal force (Fc) is the red arrow, the normal force (Fn) the blue one and the orange is the gravity and it's components. Taking the blue, red, and the opposite orange arrows: the centrifugal force in this is then the resultant force of both the gravity component and the normal force: Fc = Fzc + Fn where because gravity is 'down' and the other two are 'up' we get Fc = -Fzc + Fn, rewritten we'd get Fn = Fc - -Fzc = Fc+ Fzc.

    Would that be the correct interpretation?
     
  9. Nov 11, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
     
  10. Nov 11, 2015 #9
    Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Centrifugal and centripetal forces on a half pipe
Loading...