Physics
posted by rufy .
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 uniformdensity 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.

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

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