The discussion focuses on calculating the number of revolutions a 5.0m diameter merry-go-round makes as it decelerates from an initial period of 4.0 seconds to a complete stop over 20 seconds. The initial linear speed is calculated to be 3.9 m/s, leading to an initial angular speed of 1.56 rad/s. The average speed during deceleration is determined to be half of the initial speed, which is essential for calculating the total angle traveled. The final answer is confirmed to be 24.8 revolutions.
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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