The merry-go-round above slows down and comes to a stop in Δt = 6.00 s. What is the average angular acceleration during this time

College Physics
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Chapter7: Rotational Motion And Gravitation
Section: Chapter Questions
Problem 2P: A bicycle tire is spinning clockwise at 2.50 rad/s. During a time period t = 1.25 s, the tire is...
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5. The merry-go-round above slows down and comes to a stop in Δt = 6.00 s. What is the average angular acceleration during this time?

2. A merry-go-round goes around one complete rotation in time T= 9.00 s. What is the angular velocity, o, and
the frequency, f, of the merry go round?
Time taken to complete one revolution ; T=9 sec
As we know°, time required to complete one revolution
called as time period.
Hence time period T=9 sec
angular frequency is given by relation;
W= 2入
W = 0.698 rad/s Answer
frequency (f) given as
0.698
こ
ニ
2て
f = 0.111 Hz
Answer
Transcribed Image Text:2. A merry-go-round goes around one complete rotation in time T= 9.00 s. What is the angular velocity, o, and the frequency, f, of the merry go round? Time taken to complete one revolution ; T=9 sec As we know°, time required to complete one revolution called as time period. Hence time period T=9 sec angular frequency is given by relation; W= 2入 W = 0.698 rad/s Answer frequency (f) given as 0.698 こ ニ 2て f = 0.111 Hz Answer
3. Calculate the linear velocity, v, of horse on that merry go round that is located a distance r= 2.50 m from the
center.
Vニrd
r- distunce from center
W-Angular Speed.
Y= 2.50m
W= 0.698 rad/s
V= (2.50m) ×
V= 1.745 m/s
Linear velocity, v= 1.745 m/s
(0.698 rad/s)
Transcribed Image Text:3. Calculate the linear velocity, v, of horse on that merry go round that is located a distance r= 2.50 m from the center. Vニrd r- distunce from center W-Angular Speed. Y= 2.50m W= 0.698 rad/s V= (2.50m) × V= 1.745 m/s Linear velocity, v= 1.745 m/s (0.698 rad/s)
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