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

Step-by-step 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 * r`

Putting the given values into the above formula as follows.

`a_(tangential) = \alpha * r`

=
`11.6 rad/s^(2) * 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)`.