A flywheel with a radius of 0.300m starts from rest and accelerates with a const

Question

A flywheel with a radius of 0.300m starts from rest and accelerates with a constant angular acceleration of 0.600rad/s2 .

Part A

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start.

Enter your answers separated with commas.

Part B

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 60.0?.

Enter your answers separated with commas.

Part C

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 120?.

Enter your answers separated with commas.

Explanation / Answer

? = 0.5 (0.600 rad/s^2) t^2 = 0.180 t^2 rad <---- radial angle of point position on rim versus time
? = d?/dt = 2(0.180) t = 0.360 t <---- angular speed versus time of point on rim
? = 0.600 rad/s^2 <---- constant angular acceleration of point on rim
r = 0.300 m

a.) at t=0
tangential acceleration = ? r = (0.600 rad/s^2)(0.300 m) = 0.180 m/s^2
radial acceleration = v^2 / r = (? r)^2 / r = r ?^2 = (0.300 m)[0.360 (0)]^2 = 0 m/s^2
resultant acceleration = SQRT[(tangential acceleration)^2 + (radial acceleration)^2] = 0.180 m/s^2

b.) ? = 60 degrees = ? / 3 radians = 1.047 rad
? = 0.180 t^2 rad = 1.047 rad
t = 2.41 s
tangential acceleration = ? r = (0.600 rad/s^2)(0.300 m) = 0.180 m/s^2
radial acceleration = v^2 / r = (? r)^2 / r = r ?^2 = (0.300 m)[0.360 (2.41 s)]^2 = 0.226 m/s^2
resultant acceleration = SQRT[(tangential acceleration)^2 + (radial acceleration)^2]
= SQRT[(0.180)^2 + (0.226)^2] = 0.289 m/s^2

c.) ? = 120 degrees = 2? / 3 radians = 2.094 rad
? = 0.180 t^2 rad = 2.094 rad
t = 3.41 s
tangential acceleration = ? r = (0.600 rad/s^2)(0.300 m) = 0.180 m/s^2
radial acceleration = v^2 / r = (? r)^2 / r = r ?^2 = (0.300 m)[0.360 (3.41 s)]^2 = 0.452 m/s^2
resultant acceleration = SQRT[(tangential acceleration)^2 + (radial acceleration)^2]
= SQRT[(0.180)^2 + (0.452)^2] = 0.487 m/s^2

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