Question

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 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?

Answer 1

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    $\alpha = 11.6 rad/s^{2}$

Now, we will calculate the tangential acceleration as follows.

          $a_{tangential} = \alpha \times r$

Putting the given values into the above formula as follows.

         $a_{tangential} = \alpha \times r$

                      = $11.6 rad/s^{2} \times 3.25 cm$

                      = 37.7 $rad cm/s^{2}$

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 $rad cm/s^{2}$.