A gyroscope flywheel of radius $\mathbf{2}\mathbf{.}\mathbf{83}\mathbf{ }\mathbf{\text{cm}}$is accelerated from rest at$\mathbf{14}\mathbf{.}\mathbf{2}\mathbf{}\raisebox{1ex}{${\displaystyle \mathbf{\text{rad}}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{s}}^{\mathbf{3}}$}\right.$ until its angular speed is $\mathbf{2760}\raisebox{1ex}{${\displaystyle \mathbf{\text{rev}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{min}}}$}\right.$.

(a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process?

(b) What is the radial acceleration of this point when the flywheel is spinning at full speed?

(c) Through what distance does a point on the rim move during the spin-up?

1. The radius of flywheel r is, $2.83 \text{cm}$.
2. The angular acceleration of flywheel$\alpha$is,$14.2\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.

1. The angular speed of flywheel $\omega$ is, $2760\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.

By using the formulas for tangential acceleration $a_t$, radial acceleration$a_r$ , angular displacement $\theta$, and distance travelled$s$ , we can find thetangential acceleration and radial acceleration of a point on the rim oftheflywheel duringthespin-up process and also the distance respectively and given as.

1. Tangential acceleration, $a_t=\alpha r$
2. Radial acceleration is$a_r={\omega }^{2}r$
3. Angular displacement is$\theta =\frac{{\omega }^2}{2\alpha }$
4. The distance travelled is$s=r\theta$

We know that the tangential acceleration is given by

$a_t=\alpha r$

Substitute all the value in the above equation.

$\begin{array}{rcl}{a}_t& =& 14.2\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\mathit{\text{2×}}\mathit{2}\mathit{.}\mathit{83}\mathit{ }\mathit{\text{cm}}}}$}\right.\\ & =& 40.2\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

Hence the tangential acceleration is, $40.2\raisebox{1ex}{${\displaystyle \text{cm}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.

Step 3: (b) Calculation for the radial acceleration of this point when the flywheel is spinning at full speed

In $\text{rad/s}$, the angular velocity is given by

$\begin{array}{rcl}\omega & =& 2760\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min×}\left(\frac{2\pi }{60}\right)}$}\right.\\ & =& 289\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

So, the radial acceleration is given by

$a_r={\omega }^{2}r$

Substitute all the value in the above equation.

$\begin{array}{rcl}{a}_r& =& {\left(289\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.\right)}^2\times 0.0283 \text{m}\\ & =& 2.36\times {10}^{3}m/s^2\end{array}$

Hence the radial acceleration is, $2.36\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.

Step 3: (c) Calculation for the distance through which point on the rim move during the spin-up?

The angular displacement is

$\theta =\frac{{\omega }^2}{2\alpha }$

Substitute all the value in the above equation.

$\begin{array}{rcl}\theta & =& \frac{{\left({\displaystyle 289\text{rad/s}}\right)}^2}{{\displaystyle 2\times 14.2\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.}}\\ & =& 2.94\times {10}^3 \text{rad}\\ & & \end{array}$

Thus, the distance travelled is

$s=r\theta$

Substitute all the value in the above equation.

$\begin{array}{rcl}s& =& 0.0283 \text{m}\times 2.94\times {10}^3 \text{rad}\\ & =& 83.2 \text{m}\\ & & \end{array}$

Hence the distance is, $83.2 \text{m}$.