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 on TOTAL kinetic energy

  1. Apr 9, 2004 #1
    Can someone help me with this problem:

    A cylindrical wooden log rolls (without slipping) down a smal ramp. Tha log has radius of 15 cm, a length of 2.0m and a mass of 85kg.
    What was the log's TOTAL kinetic energy at the bottom of the ramp? (Recall that Icylinder = (1/2)MR^2) .

    The total angular velocity i got was 200 rad/s.

    I don't know if the angular velocity will help solve the problem.

    THANKS!!
     
  2. jcsd
  3. Apr 9, 2004 #2
    Not enough information. How high is the log on the ramp?

    cookiemonster
     
  4. Apr 9, 2004 #3
    Well, that is the next question I have to solve. That's all the information. Is there a way of solving this problem if the height is not given?
     
  5. Apr 9, 2004 #4
    No. The energy is determined by the height, since the only energy the log has at first is gravitational potential energy, which depends on height.

    cookiemonster
     
  6. Apr 9, 2004 #5
    The previous question states that the log rolls at a speed of 3.0m/s when it reaches the bottom of the ramp.
     
  7. Apr 9, 2004 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Well now you have enough information!

    The total KE is the sum of (1) translational KE of the center of mass (1/2MV2), and (2) rotational KE about center of mass(1/2Iω2). Since the log rolls without slipping, V =ωR.
     
  8. Apr 9, 2004 #7
    Thanks Doc Al!
     
  9. Apr 9, 2004 #8
    Then can I find the height using this equation:

    h = v^2/2g ?
     
  10. Apr 9, 2004 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Find the height using energy conservation. The initial gravitational PE at the top gets transformed into the total KE at the bottom.
     
  11. Apr 9, 2004 #10
    No because the log also has rotational KE (1/2Iω2), not just translational KE.
     
  12. Apr 10, 2004 #11
    The answer i got is: 0.62m. Can you verify if that is correct?
     
  13. Apr 10, 2004 #12
    I get a total kinetic energy of 573.75J at the bottom and a height of 6.75/g. So if you take g to be 10.9m/s2 it's right. :wink:
     
  14. Apr 10, 2004 #13
    how did you get 6.75 as the height?
     
  15. Apr 10, 2004 #14

    Doc Al

    User Avatar

    Staff: Mentor

    I get a (slightly) different answer. Maybe we should average them all together? :wink:

    Show us the equations you used, what numbers you assumed, and how you calculated the answer. (Show it algebraically first, then with numbers plugged in.)
     
  16. Apr 10, 2004 #15
    I got:

    [tex]mgh = 573.75J[/tex]

    So:

    [tex]h = \frac{573.75J}{mg} = \frac{6.75}{g}[/tex]

    My equation:

    [tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2 = \frac{1}{2}mv^2 + \frac{1}{4}mr^2\omega ^2 = \frac{3}{4}mv^2[/tex]

    [tex]h = \frac{3v^2}{4g}[/tex]
     
    Last edited: Apr 10, 2004
  17. Apr 10, 2004 #16

    Doc Al

    User Avatar

    Staff: Mentor

    I agree, of course. (I misread your answer, Chen, like psruler did. :rolleyes: )
     
  18. Apr 10, 2004 #17
    I have a question now. Let's say it is a sphere now, instead of a cylinder. The equation would be:

    [tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2 = \frac{1}{2}mv^2 + \frac{1}{5}mr^2\omega ^2 = \frac{7}{10}mv^2[/tex]

    [tex]v^2 = \frac{10gh}{7}[/tex]

    But now let's take that sphere and reduce its radius until it tends to zero (so we find ourselves with a "point" particle). Obviously the mass and the radius of the object do not affect the speed it would reach at the bottom of the hill, as that only depends on the height of it. So on one hand, we know that for point particles there is no such thing as rotational kinetic energy, so the speed squared would be 2gh. However, if we do count the rotational KE we find the speed squrared to be of 10gh/7. So which answer is the correct one? And where's the limit between a point particle and a sphere? i.e when in the process of reducing the object's radius will the speed change from one expression to the other?
     
    Last edited: Apr 10, 2004
  19. Apr 10, 2004 #18

    Doc Al

    User Avatar

    Staff: Mentor

    An excellent question!

    This is an excellent question, Chen! (I think you have the cases mixed, though.) I think when you reach that limit, the normal analysis of static friction causing rotation will break down. Surface effects will dominate.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help on TOTAL kinetic energy
  1. Total Kinetic Energy (Replies: 1)

  2. Total Kinetic Energy (Replies: 1)

  3. Total kinetic Energy (Replies: 1)

Loading...