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A gyroscope flywheel of radius 2.12 cm is accelerated from rest at 16.2 rad/s2 until its angular speed is 1820 rev/min. (a) what is the tangential acceleration of a point on the rim of the flywheel during this spin-up process

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The tangental acceleration of the flywheel of  radius 2.12 cm with a final angular velocity of 1820 rev/min and an angular acceleration of 16.2 rad/s is 0.34 m/s².

What is tangental acceleration?

This is the rate of change of tangental velocity of an object in circular motion.

To calculate the tangental acceleration, we use the formula below.

Formula:

  • a = αr............. Equation 1

Where:

  • a = Tangental acceleration
  • r = radius of the flywheel
  • α = Angular acceleration of the flywheel.

From the question,

Given:

  • r = 2.12 cm = 0.0212 m
  • α = 16.2 rad/s²

Subsitute these values into equation 1

  • a = (0.0212×16.2)
  • a = 0.34 m/s²

Hence, The tangental acceleration of the flywheel of  radius 2.12 cm with a final angular velocity of 1820 rev/min and an angular acceleration of 16.2 rad/s is 0.34 m/s².

Learn more about tangental acceleration here: https://brainly.com/question/11476496

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