Question

PHYSICS QUESTION - CAN ANYONE HELP NOW?

Alex is pushing his 20.0 kg little brother on a merry-go-round with a radius of 2.0 meters. If his little brother is on the edge of the merry-go-round and has an angular momentum of 360 kg m2/s, what is his speed?

Answer 1

Givens: m=20kg r=2.0m L=360 kg m^2/s

Unknown: Speed . (But there are two "types of speed" which are angular speed and linear speed and ARE DIFFERENT)

Formulas: L=Iω Angular momentum = inertia x angular velocity (L=Iω)

I=mr^2; ω=v/r

Therefore, L=r^2m * v/r = rmv

L=rmv

360=20*2*v

360/(20*2) =v

360/40= v

v=9m/s (linear tangent speed)

Angular velocity ω= v/r

ω= 9/2

ω= 4.5rads/s (angular speed)

In other words, both 9 and 4.5 should be right if explained correctly.

Answer 2

It's velocity when it strikes the ground is. D. 232.9 kg.m/s.

I hope this helps!!!