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Rotational Kinematics - Cylinder down a slope.

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A certain non-uniform but cylindrically symmetric cylinder has mass 9 kg, radius 1.2 m, and moment of inertia about the center of mass 7.6 kg m2. It rolls without slipping down a rough 20° incline.

    What is the acceleration of the cylinder's center-of-mass?

    2. Relevant equations

    kinematics and the rotational equivalents

    3. The attempt at a solution

    The hints I was given was to find/use 3 equations. I believe the equations I found are wrong. I seem to have some difficulty with kinematics, and now especially this rotational kinematics.

    I found:
    m*g*sin(20) - F(friction) = m*a
    T(orque) = I * alpha
    a = alpha * r

    Any help to whether these are the useful equations, and how I may go about using these? Thx. :)
  2. jcsd
  3. Nov 2, 2008 #2


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    Staff: Mentor

    What is the relationship between torque on the cylinder and the frictional force at the circumference?
  4. Nov 2, 2008 #3
    Are they the same?

    b/c the frictional force is a force not coming out from the CM; therefore it helps it spin? I believe it is the only other force, so they must be the same?
    Last edited: Nov 2, 2008
  5. Nov 2, 2008 #4
    I also tried using the gravitational force as the Torque.
    I used m*g*sin(20)=I*alpha
    m*g*sin(20)/I=alpha and alpha*R = a
    so... m*g*sin(20)*R/I=a,
    but this didn't work either.

    I just solved this problem.
    Last edited: Nov 3, 2008
  6. Nov 20, 2008 #5
    Use these equations:
    mgsin(20) - F(friction) = ma
    I(alpha) = F(friction)*r

    The only force that applies a torque is friction, because the normal force and gravity both act on the center of mass.
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