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
Dec 28, 2016

Answer:

#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#