(a) Assuming the platform is a solid disk with radius 65 cm and a uniform mass d

Question

(a) Assuming the platform is a solid disk with radius 65 cm and a uniform mass density, what is the mass of the platform?
_____________________________kg
(b) What is his new rate of rotation?
_____________________________rev/s
(c) The student becomes nervous and drops the two dumbbells so that they fall away from the platform onto the floor. What is the student's rate of rotation now? Assume the dumbbells drop onto the platform and then bounce off radially outward.
_____________________________rev/s

A student stands on a platform that is free to rotate and holds two dumbbells each at a distance of 65 cm from his central axis (see figure below). Another student gives him a push and starts the system of student, dumbbells, and platform rotating at 0.5 rev/s. The student on the platform then pulls the dumbbells in close to his chest so that they are each 22 cm from his central axis. Each dumbbell has a mass of 4.5 kg and the rotational inertia of the student, platform, and dumbbells is initially (with arms extended)12.2 kgm^2 Model each arm as a uniform rod of mass 3 kg with one end at the central axis; the length of the arm is initially 65 cm and then is reduced to 35 cm (with the elbows sticking out ). Also, model the student's trunk (body minus the arms) as a uniform solid cylinder of mass 60 kg and radius 16 cm. (a) Assuming the platform is a solid disk with radius 65 cm and a uniform mass density, what is the mass of the platform? _____________________________kg (b) What is his new rate of rotation? _____________________________rev/s (c) The student becomes nervous and drops the two dumbbells so that they fall away from the platform onto the floor. What is the student's rate of rotation now? Assume the dumbbells drop onto the platform and then bounce off radially outward. _____________________________rev/s

Explanation / Answer

The starting L value is 1.2. No need to convert to radians if the final result is in revs/second. I calculated the inertias using the same method as you.
Rod
I=1/3(3kg)(.65^2m)=.4225 kg*m^ X 2= .845 kg*m^2
Weight
I=(1kg)(.65m)^2=.1452kg*m^2 X 2= .845 kg*m^2

Those add up to 1.69. The missing piece here you forgot was the intertia of the platform. We can find this value by taking the total intertia, which is 2.4 - 1.69 = .71.

For the second part... the key is to calculate the moment of intertias...
I=1/3(3kg)(.22^2m)=.0484 kg*m^ X 2= .0968 kg*m^2
For weights
I=(1kg)(.22m)^2=.0484kg*m^2 X 2= .0968 kg*m^2
So total inertia of those two are .1936 kg*m^2. But we need to account for the platform again. Here is where I'm not entirely sure this is correct, but I said the specific moment of inertia of the platform would not change, because the student didn't move her feet, so the mass is the same thats touching the platform. :P Therefore it is still at .71.
So total inertia after = .1936 + .71 = .9036.

So L = 1.2
So L final also = 1.2
So we set L initial = IW (final)
W (final) = L initial / Inertia final

W = 1.2 / .9036 = 1.328 rev/s.

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