A 10 kg block is attached to the axis of a solid sphere (mass 4kg, radius 0.5, m

Question

A 10 kg block is attached to the axis of a solid sphere (mass 4kg, radius 0.5, moment of inertia I=2/5MR^2) by a light inestensible string that passes over a pulley (mass 2kg, radius 0.2, moment of inertia I=1/2MR^2). The coefficient of kinetic friction between the block and the plane is 0.05. The system is realesed from rest. the sphere roll withour slipping.

At the moment the block has descendent 2 meters down the incline, calculate:

the force of friction of the sphere

The coefficient of friction between the sphere and the plane.

SECTION II FREE RESPONSE Directions: Read each question carefully and write your responses on the not be scored, You must show your work to reseive credit. ead eath question carefully and write your responies on the answer sheet provided Answers in this form wil W. Use g = 10,n/'s Figure not drawn to seale by aligh 1. A 10 kg block is altached to the asis of a sold sphere (mas 4 lg, radius 0,5 m, moment of inertial inestensible string that passes over a puley (mas 2ky radlus 02 m and moment of Inertia M kinetic friction between the block and the plane is j 003. The system is slipping. id sphere (rnasa 4 kg, ladius 0.5 m momentofinertia ,-2MP by a light The coefictent of n temine and from rest ThI here rolls without twenthe block and the plane in-oos

Explanation / Answer

2 metres down the plane

PE lost by block = 10*10*2sin(37)
Work done by friction = 0.05*10*10*cos(37)
rotational ke gained by pulley = 0.5*2*0.2^2*w^2 = 0.5*2*v^2[ w = angular velocity of the system = v/0.2]
KE gained by sphere = 0.4*4*0.5^2*(v/0.5)^2 + 0.5mv^2 = 0.4*4*v^2 + 0.5*4v^2 = work done by friction on the sphere
KE gained by block = 0.5*10v^2

Hence, 200sin(37) - 5cos(37) - v^2 - 3.6v^2 - 5v^2 = 0
9.6v^2 = 116.36
v = 3.4814 m/s

Now, the sphere reaches velocity of v in distance 2 m
so, 2*2*a = v^2
a = 3.03 m/s/s

angular acc = a/r = 6.06 rad/s/s
Now. friction = mu*mg = torque for this angular acc
mu*4*10 = 0.4*4*0.5^2 * 6.06
mu = 0.06

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