Question

An automobile engine develops 100 kW power when rotating at a speed of 1800 rpm, What force does it deliver? If the diameter of the flywheel of engine is 1 m.

Answer 1

Given:

An automobile engine develops 100 kW power when rotating at a speed of 1800 rpm.

To find:

Force delivered?

Calculation:

Angular velocity be $\omega$.

$\therefore \: \omega = 1800 \: rpm$

$= > \: \omega = 1800 \times \dfrac{2\pi}{60}$

$= > \: \omega = 180 \times \dfrac{2\pi}{6}$

$= > \: \omega =60\pi \: rad {s}^{ - 1}$

Now , linear Velocity be v :

$\therefore \: v = \omega \times r$

$= > \: v = 60\pi\times \dfrac{diameter}{2}$

$= > \: v = 60\pi\times \dfrac{1}{2}$

$= > \: v = 30\pi \: m {s}^{ - 1}$

Now , power delivered :

$\therefore \: P = F \times v$

$= > \: 100 \times {10}^{3} = F \times 30\pi$

$= > \: 10 \times {10}^{3} = F \times 3\pi$

$= > \: {10}^{4} = F \times 3\pi$

$= > \: F = \dfrac{ {10}^{4} }{3\pi}$

$= > \: F = 1061.03 \: newton$

So, force delivered is 1061.03 N.