A student stands on a platform that is free to rotate and holds two dumbbells, e

Question

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. Another student gives him a push and starts the system of student, dumbbells, and platform rotating at 0.35 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 4.00 kg and the rotational inertia of the student, platform, and dumbbells is initially 9.80 kg · m2. Model each arm as a uniform rod of mass 3.00 kg with one end at the central axis; the length of the arm is initially 65 cm and then is reduced to 22 cm. What is his new rate of rotation?

I have tried this problem a million times and can't seem to get the right answer. I got 3.9 rev/s. Thanks!

Explanation / Answer

We are given that the Initial moment of inertia of the entire system is 9.8 kg m2

That I = I(student body) + I(two arms) + I(two dumbbells)

From that we can find the I of the student body. The arms are considered as rods (I = 1/3ML2), there are two of them. The dumbbells can be considered point masses (I = MR2), there are two of them.

9.8 = I(body) + (2)(1/3)(3)(.65)2 + (2)(4)(.65)2

I body = 5.575 kgm2

With this information, we can find the new total I when the arms and dumbbells are at a closer radius

Total I = (5.575) + (2)(1/3)(3)(.22)2 + (2)(4)(.22)2

Total I = 6.059

Then, by conservation of momentum

I11 = I22

I1/I2 = 2/1

(9.8)/(6.059) = (2)/(.35)

2 = .566 rev/s

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