A uniform, 8.40-kg, spherical shell 50.0 cm in diameter has four small 2.00-kg masses attached to its outer surface and equally spaced around it. This combination is spinning about an axis running through the center of the sphere and two of the small masses (?Fig. P10.56?). What friction torque is needed to reduce its angular speed from 75.0 rpm to 50.0 rpm in 30.0 s?

Solution 60 P Step 1: Data given Mass of sphere M = 8.40 kg Diameter D = 50.0 cm Radius R = D/2 = 25 cm Mass attached outside m = 2kg 75.0 rpm Angular velocity initial 1 75.0 rpm = 60 s × 1 rev 7.85 rad/s = 7.85 rad/s 1 5.0 rpm 2 Angular velocity final =250.0 rpm = 60 s × 1 rev= 5.23 rad/s 2= 5.23 rad/s Time taken t = 30 s We need to find the friction torque on the sphere Step 2 : We shall find the moment of inertia of the sphere It is given by I = (2/3)MR + 2mR 2 Substituting values we get 2 2 I = (2/3) × 8.40 kg × (25 cm) + 2 × 2kg × (25 cm) 2 2 I = 0.3500 kg m + 0.2500 kg m 2 I = 0.600 kg m