# How many revolutions does the merry go round make as it stops?

1. Nov 26, 2008

### mags

1. The problem statement, all variables and given/known data
A 5.0m diameter merry go round is initially turning with a 4.0s period. It slows down and stops in 20 s. How many revolutions does the merry go round make as it stops?

2. Relevant equations
w=v/r
v=2(pi)r/T

3. The attempt at a solution
v=2(pi)2.5/4
=3.9 m/s

w=v/r
= 3.9/2.5

2. Nov 27, 2008

### Staff: Mentor

Re: Kinematics

This is the initial angular speed. The final angular speed is zero. What's the average speed as it slows down?

(Note that ω = 2π/T.)

3. Dec 3, 2008

### mags

Re: Kinematics

i don't understand...

4. Dec 3, 2008

### tachyontensor

Re: Kinematics

Final velocity (linear and angular) is zero because the problem says the merry-go-round stops. Try thinking of this in terms of linear kinematics, making the appropriate substitutions.

5. Dec 4, 2008

### Staff: Mentor

Re: Kinematics

You found the initial angular speed ω. Good.

What you did here was to convert from rad/s to revolutions per second; so that should be 2.48 rev/s.
To get the number of revolutions, you need to compute the angle traveled by the merry go round as it comes to rest. That's just the rotational analog to Distance = speed X time. Since the speed is not constant, you must use the average speed, not the initial speed. (The average speed is just half the initial speed.)

6. Dec 4, 2008

### mags

Re: Kinematics

thanks a bunch...solved it!!!