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Rigid body rolling along an inclined plane

  1. Jun 4, 2010 #1
    1. The problem statement, all variables and given/known data
    So rigid body (ring, cylinder, sphere), who's radius of gyration is known, rolls down an inclined plane without slipping. Find velocity as a function of height.
    m - mass of the body. y0 - initial height, g - gravitational acceleration, v - speed. I - moment of inertia, K - radius of gyration. T - torque, a - linear acceleration, alfa - angular acceleration, w - angular speed. theta - inclined plane angle with horizontal.

    2. Relevant equations
    conservation of mechanical energy:
    1/2mv2+1/2Iw2+mgy = mgy0
    I = mK2
    v = wR -> w = v/R
    T = I*alfa

    3. The attempt at a solution
    using conservation of mechanical energy i get:
    1/2mv2+1/2mK2v2/R2+mgy = mgy0
    from this velocity is:
    v2 = 2g(y0-y)/(1+(K2/R2))
    which is correct answer as my textbook says.
    However I get in trouble trying do this problem in different way - using forces. As i think, if body rolls without slipping that means that force of friction is equal to the m*g*sin(theta), only friction makes this body roll. All torques about center of mass except that of friction equal to zero, and I get:
    T = R*m*g*sin(theta)
    I*alfa = R*m*g*sin(theta)
    K2alfa = R*g*sin(theta)
    and if i use relation a = alfa*R, and multiply both sides of last equation by R, I get:
    a = R2/K2*g*sin(theta)
    to get velocity as a function of height, i rewrite a = dv/dt = dx/dt*(dv/dx) = v(dv/dx):
    vdv = R2/K2*g*sin(theta)dx
    if initial speed is zero and initial coordinate along the plane also zero, i get:
    v2/2 = R2/K2*g*sin(theta)*x
    and sin(theta)*x is just y0 - y
    so my final solution is bit different from true one:
    v2 = 2g(y0-y)*R2/K2

    So, where did I mistake?
     
  2. jcsd
  3. Jun 4, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If the force of friction equaled the component of gravity parallel to the incline, then the net force would be zero and it wouldn't move down the incline.

    You're right that friction is what makes it roll, but you'd need to figure out the friction force. Don't just assume it equals m*g*sin(theta), which it doesn't.
     
  4. Jun 4, 2010 #3
    so i should have write equation of motion like this:
    ma = mg*sin(theta) - F_fric
    now from here its really clear that if F_fric = mg*sin(theta) then a = 0.
    Then if friction makes body roll, then i can write
    torque = R*F_fric
    I*alfa = R*F_fric
    F_fric = I*alfa/R.
    yes, now i get the correct answer. Thank you, I was struggling with this all morning
     
  5. Jun 4, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You got it now. :approve:
     
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