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

[mags]

Homework Statement

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?

Homework Equations

w=v/r
v=2(pi)r/T

The Attempt at a Solution

v=2(pi)2.5/4
=3.9 m/s

w=v/r
= 3.9/2.5
=1.56 rad/s

1.56 rad/s X (1rev/2(pi)rad)=2.48 rev

how is the answer 24.8? I don't understand...Please help

 

[Doc Al]

The Attempt at a Solution

v=2(pi)2.5/4
=3.9 m/s

w=v/r
= 3.9/2.5
=1.56 rad/s

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.)

 

[mags]

i don't understand...

 

[tachyontensor]

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.

 

[Doc Al]

w=v/r
= 3.9/2.5
=1.56 rad/s

You found the initial angular speed ω. Good.

1.56 rad/s X (1rev/2(pi)rad)=2.48 rev

What you did here was to convert from rad/s to revolutions per second; so that should be 2.48 rev/s.

how is the answer 24.8? I don't understand...Please help

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.)

 

[mags]

thanks a bunch...solved it!