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A gyroscope flywheel of radius 3.42 cm is accelerated from rest at 10.2 rad/s2 until its angular speed is 2220 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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Lanuel

The tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.349 m/s.

Given the following data:

  • Radius = 3.42 cm
  • Initial angular speed = 0 rev/min
  • Final angular speed = 2220 rev/min
  • Angular acceleration = 10.2 [tex]rad/s^2[/tex]

Conversion:

Radius = 3.42 cm to meter = [tex]\frac{3.42}{100} =0.0342\;meter[/tex]

To find the tangential acceleration of a point on the rim of the flywheel during this spin-up process:

Mathematically, the tangential acceleration of an object is given by the formula:

[tex]Tangential \;acceleration = radius \times angular \;acceleration[/tex]

Substituting the given parameters into the formula, we have;

[tex]Tangential \;acceleration = 0.0342 \times 10.2[/tex]

Tangential acceleration = 0.349 m/s

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