Question

I realize this is more of a physics question, but there's nobody there and it really deals with more of a mathematical aspect of it. Help would be greatly appreciated--

A merry-go-round with negligible mass has to 60 kg riding it with a distance of 4 meters from the rotational axis. They initially rotate at 3 meters a second. What is their velocity if they move to a point 1/2 a meter from the axis?

Answer 1

0.5 meter towards or away from the axis?

Answer 2

oh nvm

Answer 3

using angular velocity I got 26.3, but I think its wrong. Do you know the answer?

Answer 4

No, I don't. It's difficult, and I got put in a physics class mid-year! :-/

Answer 5

And I realized there's a typo in the question above.. 'two 60 kg children' it should say.

Answer 6

doesnt matter what their mass is since velocity doesn't depend on mass

first determine the rotational velocity since that will be the same no matter what their position on the merry go round

w = v/r
v = 3
r = 4
w = 3/4

so their rotational velocity is 0.75 m/s

now they adjusted their radius to r = 0.5

their rectangular velocity is

v = rw (manipulation of above equation)
v = 0.5 * 0.75 = 0.375 m/s

Answer 7

Thank you! What does 'w' stand for in the equation, though?

Answer 8

w is just angular velocity, it really should be omega:

\[\omega = \frac{ v }{ r }\]

Answer 9

i used the terms "angular velocity" and "rotational velocity" synonymously

Answer 10

when you think about it, on a merry go round, the father you are from the center, the faster you seem to move. that's because you have more linear velocity (rectangular velocity). but you are still covering the same angle per unit time with respect to the merry go round. i.e., you make a complete revolution in the same time as you would if you were closer to the center or anywhere else on the merry go round. so your angular velocity is constant.

the way angular velocity and linear velocity are related is by w = v/r