A flywheel with a radius of 0.290 m starts from rest and accelerates with a cons

Question

A flywheel with a radius of 0.290 m starts from rest and accelerates with a constant angular acceleration of 0.700 rad/s^2.

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

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 degrees.

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 degrees.

Explanation / Answer

here,
radius , r = 0.29 m

accelration , a = 0.7 rad/s

magnitude of the radial acceleration of a point on its rim at the start,ac = v² / r

ac = 0^2/r

ac = 0

tangential accelration , at = r*a

at = 0.29 * 0.7

at = 0.203 m/s^2

resultant accelration , a' = sqrt(ac^2 + at^2)

a' = 0.203 m/s^2

resultant accelration is 0.203 m/s^2

when rim has turned through 60.0 degrees

because angular acceleration is constant,

tangential acceleration will be constant as well
at = r *

at = 0.29 * (0.700)

at = 0.203 m/s²

for the radial acceleration of a point on its rim

wf^2 = wi^2 + 2 * a * theta

wf^2 = 0 + 2 * (0.700) * 60 *(pi/180)

wf = 1.21 rad/s

ac = f^2 * r

ac = 1.21^2 (0.29)

ac = 0.42 m/s^2

resultant accelration , a' = sqrt(ac^2 + at^2)

a' = 0.47 m/s^2

resultant accelration is 0.47 m/s^2

when rim has turned through 120 degrees

because angular acceleration is constant,

tangential acceleration will be constant as well

at = r *

at = 0.29 * (0.700)

at = 0.203 m/s²

for the radial acceleration of a point on its rim

wf^2 = wi^2 + 2 * a * theta

wf^2 = 0 + 2 * (0.700) * 120 *(pi/180)

wf = 1.71 rad/s

ac = f^2 * r

ac = 1.71^2 (0.29)

ac = 0.85 m/s^2

resultant accelration , a' = sqrt(ac^2 + at^2)

a' = 0.87 m/s^2

resultant accelration is 0.87 m/s^2

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