6. A solid sphere with mass M = 5.0 Kg is rotating about a diameter at the rate

Question

6. A solid sphere with mass M = 5.0 Kg is rotating about a diameter at the rate of 400 rev/min. If 40.0 KJ of work, without any waste, has been used to achieve this rate from rest, then

a) calculate the moment of inertia and the radius of the sphere. (ISphere =2/5MR2, generically)

b) If it takes 1.5 min to uniformly slow down this sphere to 300 rev/min, calculate the angular acceleration of the wheel during this period.

c) If the same deceleration continues, determine the angular velocity of the wheel 1.0 min later. (1.0 min after it reaches 300 rev/min.)

d) Assume this deceleration is done by a applying a constant friction force, such as pushing a rough object (shown as A in the figure) in the belly. If ?k = 0.7 between A and sphere, calculate how hard A is pushed against the sphere.

Explanation / Answer

M = 5.0 Kg
= 400 rev/min
= 400 * (2*3.14)/60 rad/s
= 41.87 rad/s

Kinetic Energy rotational is given by = 1/2 *I^2

Initial K.E = 0
Final K.E = 1/2 * I*^2

Work done = K.E fin - K.Ein
40 * 10^3 = 1/2 * I * 41.87^2
I = 45.6 kg m^2

Moment of Inertia, I = 45.6 kg m^2

Moment of Inertia of Sphere = 2/5 * MR^2
45.6 = 2/5 * 5.0 * R^2
R = 4.77 m
Radius of the Sphere, R = 4.77 m

(b)
i = 41.87 rad/s
f = 300 rev/min = 300 * (2*3.14)/60 rad/s
f = 31.4 rad/s
t = 1.5 min = 1.5 * 60 = 90 s
f = i - t
= (41.87 - 31.4)/90
= 0.116 rad/s^2
Angular acceleration of the wheel during this period, = 0.116 rad/s^2

(c)
Here
i = 31.4 rad/s
t = 1.0 min = 60 s

f = i - *t
f = 31.4 - 0.116 * 60
f =  24.44 rad/s

(d)
Torque = I * = u*Force * R
45.6 * 0.116 = 0.7*F * 4.77
F = 1.58 N

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