It is nearing Christmas time again folks and in your dream Santa pays you a visi

Question

It is nearing Christmas time again folks and in your dream Santa pays you a visit and takes you out to a 'fair' where there is a large merry go round. It's been ages since you've been on one. So excited, you stand up on it grabbing a supporting pole (assume the pole is massless). Let us say that you have a mass of 60 kg and the merry go round is rotating at 0.80 rev/s about an axis through its center. As a first approximation, let us say that the merry go round is a uniform disk of mass 210 kgand radius 4.0 m and let's say that your shape can be approximated as that of a solid (not hollow) cylindrical rod of uniform density with a length of 1.68 m and a radius of 0.33 m.

If your center of mass is 2.57 m away from the edge of the disk, what is the total angular momentum of the disk+you system?

Please express your answer to 6 sig figs with units and include steps taken to reach it. All help is appreciated, I will provide feedback.

Explanation / Answer

Moment of inertia of disk = (1/2)*M*R^2

= (1/2)*210*4^2

= 1680 kg.m^2

moment of inertia of person about center of his center os mass = (1/2)*m*r^2

moment of inertia of person about center of the disk = (1/2)*m*r^2 + m*d^2 (uning prallel axis throrem)

= (1/2)*60*0.33^2 + 60*2.57^2

= 399.561 kg.m^2

Total moment of inertia of the system, I = 1680 + 399.561

= 2079.561 kg.m^2

angular speed, w = 0.8 rev/s

= 0.8*2*pi rad/s

= 5.026 rad/s

Total angular momentum of the system = I*w

= 2079.561*5.026

= 10453.0 kg.m^2/s <<<<<<<-----Answer

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