A sphere of mass 1.1 kg and radius 0.5 m is attached to the end of a massless ro

Question

A sphere of mass 1.1 kg and radius 0.5 m is attached to the end of a massless rod of length 3.0 m. The rod rotates about an axis that is at the opposite end of the sphere (see below). The system rotates horizontally about the axis at a constant 472 rev/min. After rotating at this angular speed in a vacuum, air resistance is introduced and provides a force 0.33 N on the sphere opposite to the direction of motion. What is the power (in W) provided by air resistance to the system 167.0 s after air resistance is introduced? (Enter the magnitude.)

____ W

EXAMPLE:

A sphere of mass 1.3 kg and radius 0.5 m is attached to the end of a massless rod of length 3.0 m. The rod rotates about an axis that is at the opposite end of the sphere (see below). The system rotates horizontally about the axis at a constant 472 rev/min. After rotating at this angular speed in a vacuum, air resistance is introduced and provides a force 0.13 N on the sphere opposite to the direction of motion. What is the power (in W) provided by air resistance to the system 187.0 s after air resistance is introduced? (Enter the magnitude.)

20.1 W

Explanation / Answer

moment of inertia of sphere is I1 = (2/5)*m*R^2 = (2/5)*1.1*0.5^2 = 0.11 kg-m^2

moment of inertia of the system is I = I1+(M*(R+l)^2) = 0.11+(1.1*(0.5+3)^2) = 13.585 kg-m^2

angular speed is w = 472 rev/min = 472*(2*3.142/60) = 49.43 rad/s

torque due to friction = T = I*alpha

But T = r*F = I*alpha

3.5*0.33 = 13.585*alpha

alpha = 0.085 rad/s^2

using

wf = w -(alpha*t) = 49.43-(0.085*167) = 35.235 rad/s

work done by friction is W = 0.5*I*(wf^2-w^2) = 0.5*13.585*(35.235^2-49.43^2) = -81.63.36 J

power is P = W/t = 8163.36/167 = 48.88 W

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