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!

Help with Past Exam Question on Rolling Disk on Hemisphere

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Hoping to outdo his physics professor, Doofus wants to dramatically demonstrate parabolic motion by throwing a cheese wheel of massm and radius r off of the top of the UW observatory, which is at a height 3R above the roof of the physics building, as shown in the diagram. Just as he is finishing his climb, he accidentally lets go of the cheese wheel at the top of the hemispherical dome of radius R. It begins rolling without slipping from rest down the edge of the dome.

    media%2Fd9c%2Fd9cbae5c-26d6-456b-93b1-81592ff0af31%2FphpblwL43.png

    a) Draw a free-body diagram of the cheese when it has rolled a distance s along the dome

    b) What is the value of s when the cheese leaves the dome

    c) Diligent is standing on the roof of the physics building and catches the cheese just as it hits the ground. How fast is the cheese moving when he catches it

    2. Relevant equations
    Conservation of Energy , s = r*a, Newton's Second Law

    3. The attempt at a solution
    a) Forces are static friction, normal force and gravity
    b) Applying conservation of energy gives v^2 = 4/3 ((R+r)(1-cos(a))g. Since the wheel falls off when N =0, a = arccos(4/7). Therefore s = r*arccos(4/7).
    c) Conservation of energy gives 0.5m (v_f^2 - v_c^2) = mg(s+3R). Therefore v_f^2 = 2g(s+3R) +4/3 (R+r)(1-4/7)g.
     
  2. jcsd
  3. Dec 5, 2015 #2
    Ooops, the change in gravitational potential energy is mg(-25/7 R).
     
  4. Dec 6, 2015 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    s is the distance traveled along the dome, and r is the radius of the cheese.
     
  5. Dec 6, 2015 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    That is correct. What is the final velocity of the cheese then? Note that the rotation of the cheese does not change after leaving the dome. Velocity means the velocity of its center of mass.
     
  6. Dec 6, 2015 #5
    It would be v_f^2 = 2g(-25/7 R) + 4/3 (R+r) (3/7)g = 2/7 (2r-23R)g.
     
  7. Dec 6, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    There are several steps I'm not following.
    Where does s+3R come from? s is a distance around a curve. What has it to do with GPE change?
    What phase of the motion does the GPE change of -25/7 mgR refer to?
    Seems to me the final answer has to be of the form v2=g[4R-(R+r)c] for some constant c.
     
  8. Dec 6, 2015 #7
    Firstly s+3R was a stupid mistake. The -25/7 mgR refers to the parabolic motion of the wheel when it falls off of the dome. It falls of the dome at a height of Rcos(a), or 4/7 R and then falls the 3R for a total change in height of 25/7 R.
     
  9. Dec 6, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Are you sure about that? That's the height of what part of the cheese relative to what? Doesn't r feature?
     
  10. Dec 6, 2015 #9
    Oh, thank you. That would only be the height of the bottom of the cheese, but we need the height of the center of mass of the cheese, so it's (R+r)cos(a). Thanks :)
     
  11. Dec 6, 2015 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Still not quite right. The mass centre does not reach the ground.
     
  12. Dec 7, 2015 #11

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The cheese is certainly smaller then the dome, so r<R. That means you get negative value for vf2.
     
  13. Dec 7, 2015 #12
    Yes, I see that the velocity ends up imaginary but am not sure how to get a positive value. Am I applying energy conservation correctly for part c)?
     
  14. Dec 7, 2015 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, you should apply conservation of energy, but you did not show your work.
     
  15. Dec 7, 2015 #14
    Ok, so for c) . The velocity when the cheese wheel falls off is v^2 = g(R+r)cos(a) from the condition that the normal force goes to zero. Since cos(a) was found to be 4/7 in part b) this is v^2 = 4/7 (R+r)g. Conservation of energy tells us that 0.5m(v_f^2 -v^2) = mg(0-h), where h is the height the cheese falls through to reach the ground. Since the cheese falls off the dome at a height of 4/7 R and then falls an additional 3R to the roof h = 25/7 R. Using the result for v^2 I get the negative answer for v_f^2.
     
  16. Dec 7, 2015 #15

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You have a sign error (and some other mistakes). Write out the energy as PE + KE both at the initial and final positions.You will find the sign error. Include also r. what is h?
     
  17. Dec 7, 2015 #16
    I found the sign error but still can't figure out what h should be.
     
  18. Dec 7, 2015 #17

    ehild

    User Avatar
    Homework Helper
    Gold Member

    At what height above the ground is the center of mass of the cheese when it leaves the dome? And at what height it is when the cheese reaches the ground?
     
  19. Dec 7, 2015 #18
    Oh, it's at height r when the cheese hits the ground and 4/7 (R+r) + 3R when it leaves the dome.
     
  20. Dec 7, 2015 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes.
     
  21. Dec 7, 2015 #20
    Ok, thank you so much for the patient help.
     
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: Help with Past Exam Question on Rolling Disk on Hemisphere
Loading...