The problem involves a merry-go-round with a diameter of 5.0m that is initially rotating with a period of 4.0 seconds and comes to a stop over a duration of 20 seconds. Participants are discussing how to calculate the number of revolutions made during this deceleration phase.
Some participants have provided calculations for initial angular speed and have noted the need to consider average speed for the total distance traveled in revolutions. There is an ongoing inquiry into the correct interpretation of the problem and the calculations involved.
There is a mention of confusion regarding the relationship between angular speed and the total number of revolutions, as well as the need to account for the average speed during deceleration. The final velocity is stated to be zero, as the merry-go-round stops.
This is the initial angular speed. The final angular speed is zero. What's the average speed as it slows down?mags said: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
You found the initial angular speed ω. Good.mags said:w=v/r
= 3.9/2.5
=1.56 rad/s
What you did here was to convert from rad/s to revolutions per second; so that should be 2.48 rev/s.1.56 rad/s X (1rev/2(pi)rad)=2.48 rev
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.)how is the answer 24.8? I don't understand...Please help