1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Block slides from rest from the top of a fixed frictionless sphere

  1. Nov 12, 2008 #1
    can anyone help me to solve this question? please and thank you :D

    A small block slides from rest from the top of a fixed frictionless sphere of
    radius R.

    Determine the vertical distance traveled by the block, x , where it loses
    contact with the surface of the sphere and its speed at this point.
  2. jcsd
  3. Nov 13, 2008 #2
    let m be the mass of the block.
    let y be the angular displacement of the block between its starting point and the point in which it loses contact with the sphere.

    we can see that cos(y) = (R-x)/R, x = R[1 - cos(y)]

    since there is conservation of mechanical energy, loss of gravitational potential energy will result in a gain in kinetic energy of the block, mgx = (1/2)mv2
    hence, v2 = 2gx = 2gR[1 - cos(y)]

    next, as the block moves down the sphere, a portion of its weight must act radially into the sphere so as to provide the required centripetal force for the motion of the block.
    hence, required centripetal force = mg*cos(y) = mv2/R
    substituting the value of v2 found previously into the equation, mg*cos(y) = 2mgR[1 - cos(y)]/ R
    hence, rearranging the terms, we get cos(y) = 2/3

    since x = R[1 - cos(y)], x = R(1 - 2/3) = R/3

    since v2 = 2gx = 2gR[1 - cos(y)], v2 = 2gR/3, v = (2gr/3)1/2

    is it correct?

    Attached Files:

    • a.pdf
      File size:
      3.1 KB
  4. Nov 13, 2008 #3
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook