Asked by Angular Momentum
A runner with mass 100 kg jogs around a circular track with circumference 1 km. If he
does one lap in 6 min, what is his angular momentum about the center of the track?
does one lap in 6 min, what is his angular momentum about the center of the track?
Answers
Answered by
drwls
In this case,
Angular Momentum = M*V*R
R = 1000 m/(2 pi) = 159.2 m
V = 1000 m/(360 s) = 2.778 m/s
Now do the MVR multiplication
Angular Momentum = M*V*R
R = 1000 m/(2 pi) = 159.2 m
V = 1000 m/(360 s) = 2.778 m/s
Now do the MVR multiplication
Answered by
john
44000kg
Answered by
john
Theradiusofthetrackis R= C =1000m=160m.Thespeedoftherunneris 2π 2π
v = d = 1000m = 2.78 ms . The angular momentum about the center is thus t 360s
L = mvR = (100kg)(2.78 m )(160m) = 44000 Js or 44000 kg m2 . ss
v = d = 1000m = 2.78 ms . The angular momentum about the center is thus t 360s
L = mvR = (100kg)(2.78 m )(160m) = 44000 Js or 44000 kg m2 . ss
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.