Question

The angular speed of a wheel increases from 600 rpm to 1200 rpm in 20 seconds.
Calculate (i) angular acceleration
(ii) angular displacement and
(iii) number of revolutions does it make in this period of time

Answer 1

take our angular displacement by formula s=∆s/r
#HopeItHepls :)

Answer 2

$\alpha =3.14\ rad/s^2$

$\theta=1884.6\ radian$

n = 300 revolution

Explanation:

Given that,

Initial speed of the wheel, $\omega_i=600\ rpm=62.83\ rad/s$

Final speed of the wheel, $\omega_f=1200\ rpm=125.66\ rad/s$

(1) The rate of change of angular velocity per unit time is called the angular acceleration of the wheel. It is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{125.66-62.83}{20}\ rad/s^2$

$\alpha =3.14\ rad/s^2$

(2) Let $\theta$ is the angular displacement. It can be calculated using second equation of motion as :

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2$

$\theta=62.83\times 20+\dfrac{1}{2}\times 3.14\times 20^2$

$\theta=1884.6\ radian$

(3) Let there are n number of revolutions it make in this period of time. It is given by :

$n=\dfrac{\theta}{2\pi}$

$n=\dfrac{1884.6}{2\pi}$

n = 299.94

or

n = 300 revolution

Learn more,

Rotational kinematics

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