A large horizontal circularplatform (M=138.1 kg, r=3.21 m) rotates about a frict

Question

A large horizontal circularplatform (M=138.1 kg, r=3.21 m) rotates about a frictionlessvertical axle. A student (m=64.3 kg) walks slowly from the rim ofthe platform toward the center. The angular velocity of thesystem is 1.10 rad/s when the student is at the rim. Find (in rad/s) when the student is 1.75 m from the center. A large horizontal circularplatform (M=138.1 kg, r=3.21 m) rotates about a frictionlessvertical axle. A student (m=64.3 kg) walks slowly from the rim ofthe platform toward the center. The angular velocity of thesystem is 1.10 rad/s when the student is at the rim. Find (in rad/s) when the student is 1.75 m from the center.

Explanation / Answer

Moment OfInertia Of Circular Platform = IP      = (1/2)*M*r^2      = (1/2)*(138.1)*(3.21)^2      = 711.498 kg*m^2 Moment Of Inertial Of Circular Platform + Student = IP+SInitial =      = IP +m*(RI)^2      = (711.498) + (64.3.3)*(3.21)^2      = 1374.05163 kg*m^2 Moment Of Inertial Of Circular Platform + Student = IP+SFinal =      = IP +m*(RF)^2      = (711.498) + (64.3)*(1.75)^2      = 908.41675 kg*m^2 We now apply Conservation of Angular Momentum: Initial Angular Momentum = Final AngularMomentum {IP+SInitial}*I = {IP+SFinal}*F (1374.05163)*(1.75) = (908.41675)*F   F =2.647 rad/s

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