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A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.4 m down a q = 29° incline. The sphere has a mass M = 3.7 kg and a radius R = 0.28 m.

What is the magnitude of the frictional force on the sphere?

--The other person did this for me but it's still saying it's wrong.

The speed acquired at the bottom is related to the height of the incline,
H = 3.4 sin 29 = 1.648 m

For a uniform-density sphere that is not slipping, conservation of energy requires that
(1/2)M V^2 + (1/2)(2/5)V^2 = M g H
V = sqrt(10/7)gH = 4.80 m/s
The acceleration rate (a) of the sphere is such that
V = sqrt(2 a X)
a = V^2/2X = 3.4 m/s^2

The angular acceleration rate is
alpha = a/R = 12.14 radian/s^2

The friction force can now be obtained from the equation relating angular acceleration to torque. The friction force F provides the torque needed to make it spin as it rolls dwn the plank.

F*R = I*alpha = (2/5)MR^2*alpha
F = (2/5)MR*alpha
= (0.4)*3.7 kg*0.18 m*12.14 s^-2
= 3.23 N

still saying it's wrong

|f| = N
3.23 NO

HELP: The frictional force provides the torque needed to give an angular acceleration. Therefore, first find the angular acceleration, then apply the rotational equivalent of Newton's 2nd Law (torque = I*a).
HELP: Since the sphere rolls without slipping, the angular acceleration is related to the translational acceleration, which can be found from kinematic relations.

  • Physics typo -

    (1/2)M V^2 + (1/2)(2/5)V^2 = M g H
    well that should be
    (1/2)M V^2 + (1/2)(2/5)MV^2 = M g H
    I think

  • Physics -

    But that is just a typo, the next line is good.

  • Physics -

    The angular acceleration rate is
    alpha = a/R = 12.14 radian/s^2

    agree so far

  • found disagreement -

    F*R = I*alpha = (2/5)MR^2*alpha
    F = (2/5)MR^2*alpha
    = (0.4)*3.7 kg* .28^2 m^2 *12.14 s^-2
    = 1.41 N

  • Wow , another typo -

    F*R = I*alpha = (2/5)MR^2*alpha
    F = (2/5)M*R*alpha (divide both sides by R)
    = (0.4)*3.7 kg* .28m *12.14 s^-2
    = 5.03 N

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