A uniform, 9.00 -kg, spherical shell 50.0 cm in diameter has four small 1.95 -kg

Question

A uniform, 9.00-kg, spherical shell 50.0 cm in diameter has four small 1.95-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 (see figure below). What friction torque is needed to reduce its angular speed from 73.5 rpm to 51.0 rpm in 28.5 s?

magnitude N A uniform, 9.00-kg, spherical shell 50.0 cm in diameter has four small 1.95-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 (see figure below). What friction torque is needed to reduce its angular speed from 73.5 rpm to 51.0 rpm in 28.5 s?

Explanation / Answer

As we know that

w = w0 + alpha*t

Therefore

(51*2pi/60) = (73.5*2pi/60) + alpha*28.5

alpha = - 0.08267 rad/sec^2

As we know

Torqu = I*alpha

So

Frictional torque = ((2/3)*9*0.25^2 + 2*1.95*0.25^2)*0.08267

= 0.05115 Nm

As it is Frictional torque

So

Direction is Opposite to Rotation

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