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!

Rotational Motion

  1. Dec 10, 2004 #1
    A 13 kg cylinder with a .54 m diameter rolls without slipping down a 30 degree incline from a height of 1.25 m.

    a. If the cylinder has I = (mr^2)/2, what will its speed be at the incline's base?

    After using the conservation of mechanical energy I got v = sqrt[(4/3)gh], which is 4.04 m/s.

    b. What is its speed if it rolls from the same height down a 60 degree incline?

    Now wouldn't the speed be the same as in part a since v is not dependent on the angle?

    My major question is why are both speeds the same if the angle changes?

    If my calculations are incorrect, please inform me. Thank you for any help.
  2. jcsd
  3. Dec 10, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor



    Since the initial PE is the same in both cases, the final speed will be the same. What does change is the acceleration down the incline. The smaller angle produces a smaller acceleration, but the distance is greater by the same factor, so the final speed remains the same.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Rotational Motion
  1. Rotative motion (Replies: 3)

  2. Rotational Motion (Replies: 5)