Problem: An object moves around a circle of radius 4.6 meters, making a revolution every 2.4 seconds. Assume that its angular velocity is constant. Starting at t = 0, when its angular position is 0 radians, what are the x and y coordinates of its position after (1/12)( 2.4) seconds, and after (5/8)( 2.4) seconds?
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Solution: Moving through a revolution, which is 2 `pi radians, in 2.4 seconds, the object will have an angular velocity of
`omega = 2 `pi radians/( 2.4 seconds) = 2.618 radians/second.
After (1/12)( 2.4) seconds, starting at 0 radians when t = 0, the angular position will be 1/12 of a complete circle, or `theta1 = (1/12)(2 `pi ) radians = `pi /6 radians.
This is a familiar angle, 30 degrees above the x axis, with sine equal to 1/2 and cosine equal to `sqrt(3) / 2. On a circle of radius 4.6 meters, the x and y coordinates will therefore be
x1 = 4.6 meters (`sqrt(3) / 2) = 3.983 meters
and
y1 = 4.6 meters (1/2) = 2.3 meters.
After (5/8)( 2.4 seconds), the angular position will be
`theta1 = (5/8)(2 `pi radians) = 5 `pi /4 radians.
This is a third-quadrant angle at 45 degrees from the x axis, with both sine and cosine equal to -`sqrt(2) / 2. On a circle of radius 4.6 meters, the x and y coordinates will therefore be
x2 = 4.6 meters (-`sqrt(2) / 2) = -.2526 meters
and
y2 = 4.6 meters (-`sqrt(2) / 2) = -.2526 meters.