Answer to Question #240503 in Mechanics | Relativity for Sachi

Question #240503

plz help!!!


A merry-go-round with 5 m radius is rotating around its vertical axis with a rate of 0.30 rev/s. A 90 kg man jump on to the merry-go-round and he lands and remains at a distance of 3 m from the axis of rotation. If the moment of inertia of the merry-go-round is 10,000

kgm2, find the new rate of rotation after the person has landed.


1
Expert's answer
2021-09-22T07:05:29-0400

Given:

r=5mr=5\:\rm m

n=30rev/sn=30\:\rm rev/s

I=10000kgm2I=10000\:\rm kg\cdot m^2

m=90  kgm=90\;\rm kg

d=3md=3\:\rm m


The law of conservation of angular momentum says

In=InIn=I'n'

I=I+md2I'=I+md^2

Hence

n=II+md2n=1000010000+903230=28rev/sn'=\frac{I}{I+md^2}n=\frac{10000}{10000+90*3^2}*30=28\:\rm rev/s

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment