The flywheel of a steam engine runs with a constant angular velocity of . When steam is shut off, the friction of the bearings and of the air stops the wheel in $\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{\text{h}}$.

(a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown?

(b) How many revolutions does the wheel make before stopping?

(c) At the instant the flywheel is turning at$\mathbf{75}\raisebox{1ex}{${\displaystyle \mathbf{\text{rev}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{min}}}$}\right.$ , what is the tangential component of the linear acceleration of a flywheel particle that is$50 \text{cm}$ from the axis of rotation?

(d) What is the magnitude of the net linear acceleration of the particle in (c)?

1. The constant angular velocity of flywheel of steam engine $\omega$ is, $150 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.
2. The air stops the wheel when t is$2.2 \text{h}$.
3. At the instant, the flywheel is rotating at angular speed$\omega$is$75 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.
4. Distance from the axis of rotation r is, $0.50 \text{m}$.

By using formulas for the angular acceleration α, the kinematic equation θ, the tangential acceleration a_t,and radial acceleration a_r, we can find theconstant angular acceleration, in revolution per minute squared, of the wheel during the slowdown, the number of revolution the wheel makes before stopping, the tangential component of the linear acceleration of a flywheel particle that is 50 cmfrom the axis of rotation, and themagnitude of the net linear acceleration of the particle in (c) respectively.

1. The angular acceleration is $\alpha =\frac{\Delta \omega }{\Delta t}$

1. The kinematic equation is $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$
2. The tangential acceleration is $a_t=\alpha r$
3. The radial acceleration is $a_r={\omega }^{2}r$

Angular acceleration is given by

$\alpha =\frac{\Delta \omega }{\Delta t}$

Substitute all the value in the above equation.

role="math" localid="1660906926211" $$\begin{gathered} \alpha =\frac{{\displaystyle 0-150\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.}}{{\displaystyle 2.2\times 60 \text{min}}} \\ \alpha =-1.14 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{min}}^{\text{2}}}$}\right. \end{gathered}$$

Hence the value of angular acceleration is, $-1.14 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{min}}^{\text{2}}}$}\right.$.

Using kinematic equation with $t=2.2\times 60 \text{min}=132 \text{min}$, we get

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

Substitute all the value in the above equation.

$$\begin{gathered} \theta =150\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.\times 132 \text{min}+\frac{1}{2}\times \left(-1.14\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${{\displaystyle \text{min}}}^2$}\right.\right)\times {\left(132 \text{min}\right)}^2 \\ =9.9\times {10}^3 \text{rev} \end{gathered}$$

Hence the number of revolution is, $9.9\times {10}^3 \text{rev}$.

Step 3: (c) Calculation for the tangential component of the linear acceleration of a flywheel particle

With $r=500 \text{mm}$, the tangential acceleration is

$a_t=\alpha r$

Substitute all the value in the above equation.

role="math" localid="1660907996924" $$\begin{gathered} a_t=\left(-1.14 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{min}}^{\text{2}}}$}\right.\right)\times \frac{2\pi \text{rad}}{1 \text{rev}}\times {\left(\frac{1 \text{min}}{60 \text{s}}\right)}^2\times 500 \text{mm} \\ =-0.99mm/{\text{s}}^{\text{2}} \end{gathered}$$

Hence the magnitude of tangential linear acceleration is, $=-0.99mm/{\text{s}}^{\text{2}}$.

The angular speed of the flywheel is

$$\begin{gathered} \omega =\left(75\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.\right)\times 2\pi \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{rev}}$}\right.\times \frac{1 \text{min}}{60 \text{s}} \\ \omega =7.85\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

With $r=0.50 \text{m}$, the radial acceleration is given by

$\begin{array}{rcl}{a}_r& =& {\omega }^{2}r\\ & =& {\left(7.85\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\times 0.50 \text{m}\\ & \approx & 31\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

which is much bigger than ${\text{a}}_t$.

Consequently, the magnitude of acceleration is given by

$\left|\overrightarrow{a}\right|=\sqrt{{a_r}^2+{a_t}^2}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\left|\overrightarrow{a}\right|& =& \sqrt{{31}^2+{\left(-0.99\right)}^2}\\ & =& 31\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & \approx & a_r\end{array}$

Hence the magnitude of linear acceleration is, $31 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.