Posted by **rufy ** on Friday, April 9, 2010 at 12:01am.

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?

- physics -
**drwls**, Friday, April 9, 2010 at 2:33am
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

- physics -
**rufy**, Friday, April 9, 2010 at 1:38pm
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.

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