What is the angular momentum of a rod with a mass of 8 kg and length of 6 m that is spinning around its center at 5 Hz?

1 Answer
Morgan
Dec 28, 2016

vecL~~754(kgm^2)/s

Explanation:

Angular momentum is given by vecL=Iomega, where I is the moment of inertia of the object, and omega is the angular velocity of the object.

The moment of inertia of a thin, rigid rod rotating about its center is given by I=1/12ML^2 and angular velocity is given by omega=2pif, where f is the frequency.

Given that f=5Hz, we can calculate the angular velocity:

omega=2pif=2pi(5s^-1)=10pi(rad)/s

Given M=8kg and L=6m, we can calculate the moment of inertia:

I=1/12ML^2=1/12(8kg)(6m)^2=24kgm^2

Using our determined values for I and omega, the angular momentum is found by:

vecL=Iomega=(10pi(rad)/s)(24kgm^2)=240pi(kgm^2)/s

=>vecL~~754(kgm^2)/s

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