Answer:
We know that the carousel does a complete rotation in 45 seconds.
Then the frequency of this carousel will be f = Â 1/45 seconds.
And the angular frequency will be 2*pi times the frequency, then we have:
angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1
Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:
S = R*W
then:
a) in this case the radius is 2.75m, then the linear speed is:
S = 2.75m*0.1396 s^-1 = 0.3839 m/s
b) in this case the radius is 1.75m, then the linear speed here is:
S = 1.75m*0.1396 s^-1 = 0.2443 m/s
(a) The linear speed of an outer horse on the carousel is 0.384 m/s.
(b) The linear speed of an inner horse on the carousel is 0.244 m/s.
Given data:
The time interval for the rotation of carousel is, t = 45 s.
The distance of the outer horse from the axis of rotation is, r = 2.75 m.
The distance of an inner horse from the axis of rotation is, r' = 1.75 m.
(a)
The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,
v = 2 π r/t
Solving as,
v = 2 π (2.75) / 45
v = 0.384 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.
(b)
Now similarly the linear speed of an inner horse is calculated as,
v' = 2 π r' / t
Solving as,
v' = 2 π (1.75) / 45
v' = 0.244 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.
Learn more about the rotational mechanics here:
https://brainly.com/question/18303026
