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Spherical Rotation question

  1. Apr 3, 2009 #1
    Q1.) A hollow, spherical shell with mass 1.70 rolls without slipping down a slope angled at 38.0.

    -Find the acceleration.
    -Find the friction force.
    -Find the minimum coefficient of friction needed to prevent slipping.

    If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.
    M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2
    = [(1/2) + (1/3)] M V^2
    The M's and R's cancel and
    gH sin 38 = (5/6) V^2
    V = sqrt (1.2 g H sin 38) = sqrt (2 a H)
    where a is the acceleration
    a = 0.6 g sin 38

    = 1.74 which is wrong



    Use that acceleration and Newton's second law t compute the actual friction force, F.
    M g sin 38 - F = M a
    Mg sin 38 - M*0.6 g sin 38 = F
    F = 0.4 M g sin 38

    =1.975 which is wrong

    To provide this amount of friction, the static coefficient of friction mu,s must equal or exceed a value given by
    M g cos 38 * mu,s = 0.4 M g sin 38
    mu,s = 0.4 tan 38

    =.124 which is also wrong.

    Can someone please check over what i did and see if i miscalculated or something.
     
  2. jcsd
  3. Apr 3, 2009 #2

    LowlyPion

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    6/5*g*x*sin38 = 2*a*x

    a = .6*g*sin38 = .6*9.8*.616 = 3.63 m/s2
     
  4. Apr 3, 2009 #3
    ok how does the sin(38) = .616?

    i keep getting .296369

    It is the right answer though
     
  5. Apr 3, 2009 #4
    You're evaluating the sine in units of radians. Switch to degrees (or multiply 38 with pi/180 before evaluating the sine).
     
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