1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Hemisphere on a ramp

  1. Jan 26, 2010 #1
    [/tex]1. The problem statement, all variables and given/known data
    http://emweb.unl.edu/negahban/em223/sexam3/sexam3.htm" [Broken]

    number 3...the one about the semicylinder on a ramp

    2. Relevant equations
    [tex]\Sigma[/tex] F_x = 0
    [tex]\Sigma[/tex] F_y = 0
    [tex]\Sigma[/tex] M=0
    f = [tex]\mu[/tex] * N

    3. The attempt at a solution
    I got the first part to be 16.7 degrees. However, I can't find \phi I just need to be pointed in the right direction...
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 26, 2010 #2
    If you were to draw a line from G to the point of contact between the cylinder and ramp, phi occurs when that line coincides with the direction of mg.
  4. Jan 26, 2010 #3
    So then I know that the distance between G and the ramp is r*(1-4/(3*pi)), but I don't know how that helps me solve the problem...
  5. Jan 26, 2010 #4
    I wouldn't say that would be the distance...that would be the case when the line from G to contact falls on a radial line, which wouldn't be the same case as when the line from G to contact falls on the vector mg.
  6. Jan 26, 2010 #5
    right...that makes sense. I'm still having a hard time picturing how this fact will help me solve the problem...
    isn't the distance between G and the contact surface going to be in the same direction as mg, otherwise the semi-cylinder would try and move back to the place where they are in the same direction (because g will cause a torque???) sorry if that doesn't make sense...
  7. Jan 26, 2010 #6
    Yup...that is what is going on...and the line covering that distance for this particular case does not necessarily coincide with a radial line (by radial line, I'm referring to the line from the center of what would have been the complete cylinder to the point of contact).
    Last edited: Jan 26, 2010
  8. Jan 26, 2010 #7
    So right before it is going to tip, torque = zero...maybe?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Hemisphere ramp Date
Ball kicked off of hemisphere Feb 25, 2018
Finding dA of a hemisphere Feb 3, 2018
Marble rolling on ramp harmonic motion Jan 21, 2018
Image distance from the hemisphere lens Dec 22, 2017