A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes
4.3 revolutions each minute.
a. What is the period of the motion?
b. What is the speed of the plane?
c. What is the acceleration of the model plane?
To answer these questions, we need to understand some basic concepts of circular motion.
a. To find the period of the motion, we can use the formula:
Period (T) = 1 / Frequency
Since the plane makes 4.3 revolutions each minute, the frequency is 4.3 revolutions/min. Plugging this value into the formula, we get:
T = 1 / 4.3 revolutions/min = 0.2326 min/revolution
Therefore, the period of the motion is approximately 0.2326 minutes per revolution.
b. The speed of the plane can be calculated using the formula:
Speed (v) = 2πr / T
The plane is flying in a circular path on the end of a 23-meter line, so the radius (r) of the circular path is 23 meters. Plugging this value along with the period we calculated (T = 0.2326 min/revolution) into the formula, we get:
v = 2π(23 meters) / 0.2326 min/revolution
Simplifying the equation, we find:
v ≈ 598.2 meters/min
Therefore, the speed of the plane is approximately 598.2 meters per minute.
c. The acceleration of the model plane can be determined using the formula:
Centripetal Acceleration (a) = v^2 / r
Using the speed we calculated (v ≈ 598.2 meters/min) and the radius of the circular path (r = 23 meters), we can plug these values into the formula:
a = (598.2 meters/min)^2 / 23 meters
Simplifying the equation, we find:
a ≈ 15,453.8 meters/min^2
Therefore, the acceleration of the model plane is approximately 15,453.8 meters per minute squared.