Home/Class 11/Physics/

# Question and Answer

## QuestionPhysicsClass 11

The motor of an engine is erotating about its axis with an angular velocity of 100 rev/minute. It comes to rest in 15 s, after being switched off. Assumgn cnstant angular decelertion, calculate the number of revolutions made by it before coming to rest.

## Answer

The initila ngular velocity =100 rev/minute =(10pi/)rad/s$${F}{i}\frac{{N}}{{A}}{l}{a}{n}{g}\underline{{a}}{r}{v}{e}{l}{o}{c}{i}{t}{y}={0}{T}{i}{m}{e}\int{e}{r}{v}{a}{l}={15}{s}\lt{b}{r}.{L}{e}{t}{a}{n}{g}\underline{{a}}{r}{a}{\mathcal{{e}}}\le{r}{a}{t}{i}{o}{n}{b}{e}$$alpha$$.{U}{\sin{{g}}}{t}{h}{e}{e}{q}{u}{a}{t}{i}{o}{n}$$omega=omega_0+alphtat$${w}{e}{o}{b}{t}{a}\in\alpha={\left(-{2}\frac{\pi}{{9}}\right)}{r}{a}\frac{{d}}{{s}^{{2}}}$$
The angle rotated by teh motor during this motion is
$$\theta=\omega_{{0}}{t}+\frac{{1}}{{2}}\alpha{t}^{{2}}$$
$$={\left(\frac{{{10}\pi}}{{3}}\frac{{{r}{a}{d}}}{{s}}\right)}{\left({15}{s}\right)}-\frac{{1}}{{2}}{\left(\frac{{{2}\pi}}{{9}}\frac{{{r}{a}{d}}}{{s}}\right)}{\left({15}{s}\right)}^{{2}}$$
Hence the motor rotates thrugh 12.5 revolutions before coming to rest.
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
Correct20
Incorrect0
Still Have Question?

Load More