Chapter 8, Problem 75GP

(a)

To determine

To Find: The total energy stored in the flywheel.

Answer to Problem 75GP

  $\mathrm{E}=1.7\times {10}^8\text{ J}$

Explanation of Solution

Given data:

Mass of the flywheel, ${\mathrm{M}}_{\mathrm{f}}=\text{ 24}0\text{ kg}$

Mass of the car, ${\mathrm{M}}_{\mathrm{c}} =\text{14}00\text{ kg}$

Diameter of flywheel, $\mathrm{d} =\text{1}.\text{5}0\text{ m}$

Radius of the flywheel, $\mathrm{r}=\mathrm{d}/\text{2}=0.\text{75 m}$

Speed of the car, ${\mathrm{v}}_{\text{c}} =\text{95}\text{kmph}=\text{26}.\text{4 m}/\text{s}$

Frictional force, ${\text{F}}_{\text{f}} =\text{45}0\text{ N}$ .

The distance travelled, $\mathrm{s} =\text{35}0\text{ km}=\text{35}0\times \text{1}{0}^{\text{3}}\text{m}$

Formula used:

Write the expression for the total energy stored in the flywheel.

  $\mathrm{E}={\mathrm{F}}_{\mathrm{f}}\times \mathrm{s}+\frac{1}{2}{\mathrm{m}}_{\mathrm{c}}{\mathrm{v}}_{\mathrm{c}}{}^2$

Here, ${\text{F}}_{\text{f}}$ is the Frictional force.

  $\mathrm{s}$ is the distance travelled,

  ${\mathrm{m}}_{\mathrm{c}}$ is the mass of the car,

  ${\mathrm{v}}_{\text{c}}$ is the speed of the car.

Calculation:

The total energy stored in the flywheel is E is given as,

  $\mathrm{E}={\mathrm{F}}_{\mathrm{f}}\times \mathrm{s}+20\times \frac{1}{2}{\mathrm{M}}_{\mathrm{c}}{\mathrm{v}}_{\mathrm{c}}{}^2$

  $\mathrm{E}=450\times 350\times {10}^3+20\times \frac{1}{2}\times 1400\times {(26.4)}^2$ J

  $\begin{array}{l}\mathrm{E}=1.67\times {10}^8\text{ J}\\ \mathrm{E}=1.7\times {10}^8\text{ J}\end{array}$

Conclusion:

The total energy stored in the flywheel to uphill and downhill is $\mathrm{E}=1.7\times {10}^8\text{ J}$ .

(b)

To determine

To Calculate: The angular velocity of the flywheel when it has a full energy charge.

Answer to Problem 75GP

  $\mathrm{\omega }=2244\text{ rad/s}$

Explanation of Solution

Given data:

Mass of the flywheel, ${\mathrm{M}}_{\mathrm{f}}=\text{24}0\text{ kg}$

Mass of the car, ${\mathrm{M}}_{\mathrm{c}} =\text{14}00\text{ kg}$

Diameter of flywheel, $\mathrm{d} =\text{1}.\text{5}0\text{ m}$

Radius of the flywheel, $\mathrm{r}=\mathrm{d}/\text{2}=0.\text{75 m}$

Speed of the car, ${\mathrm{v}}_{\text{c}} =\text{95km}\text{ph}=\text{26}.\text{4 m}/\text{s}$

Frictional force, ${\text{F}}_{\text{f}} =\text{45}0\text{ N}$ .

The distance travelled, $\mathrm{s} =\text{35}0\text{ km}=\text{35}0\times \text{1}{0}^{\text{3}}\text{m}$

Formula used:

  $\mathrm{I}=\frac{1}{2}{\mathrm{M}}_{\mathrm{f}}{\mathrm{R}}_{\mathrm{f}}^2$

  $\mathrm{E}=\frac{1}{2}\mathrm{I}{\mathrm{\omega }}^2$

Calculation:

The moment of inertia of the flywheel (cylindrical)

  $\mathrm{I}=\frac{1}{2}{\mathrm{M}}_{\mathrm{f}}{\mathrm{R}}_{\mathrm{f}}^2$

  $\mathrm{I}=\frac{1}{2}\times 240\times {(0.75)}^2$

  $\text{kg}\cdot {\text{m}}^{\text{2}}$ .

  $\mathrm{I}=67.5\text{ kg}\cdot {\text{m}}^2$

Now, the angular velocity of the flywheel can be determined by the energy of the flywheel equation.

  $\mathrm{E}=\frac{1}{2}\mathrm{I}{\mathrm{\omega }}^2$ J

  $\mathrm{\omega }=\sqrt{\frac{2\mathrm{E}}{\mathrm{I}}}\text{ rad/s}$

  $\mathrm{\omega }=\sqrt{\frac{2\times 1.7\times {10}^8}{67.5}}\text{ rad/s}$

  $\mathrm{\omega }=2244\text{ rad/s}$

Conclusion:

The angular velocity of the flywheel is $\text{2244 rad}/\text{s}.$

(c)

To determine

To Calculate: The time period for which $\text{15}0\text{ hp}$ motor gives the flywheel a full energy charge before the trip.

Answer to Problem 75GP

  $\mathrm{t}=25.23\text{ min}\text{.}$

Explanation of Solution

Given data:

Energy stored in fly wheel= $\text{1}.\text{7}\times \text{1}{0}^{\text{8}}\text{J}$

Motor horse power= $\text{15}0\text{ hp}$ .

Formula used:

  $\mathrm{t}=\frac{\text{Energy stored in flywheel}}{\text{motor HP}}$

Calculation:

  $\mathrm{t}=\frac{\text{1}\text{.7}\times {\text{10}}^8}{150\times 746}\text{s}$

  $\text{1 hp}=\text{746 W=746 J}/\text{s}.$

  $\mathrm{t}=1519.2\text{ s}$ .

  $\mathrm{t}=25.32\text{ min}$

Conclusion:

The time is $25.32\text{ min}$ , for which 150 hp motor gives the flywheel a full energy charge before the trip.