Question #280736

A 4.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s. Its angular momentum about the center of the circle is:


a: 12 kg.m2 /s


b. 24 kg.m2 /s


c. 48 kg.m2 /s


d. 6 kg.m2/s


1
Expert's answer
2021-12-20T10:28:04-0500

Given:

m=4.0kgm=4.0\:\rm kg

r=0.50mr=0.50\:\rm m

ω=12rad/s\omega=12\:\rm rad/s


The angular momentum is given by

L=Iω=mr2ωL=I\omega=mr^2\omega

L=4.00.50212=12kgm2/sL=4.0*0.50^2*12=12\:\rm kg\cdot m^2/s


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