Question

When the gear rotates 20 revolutions, it achieves an angular velocity of v = 30 rad>s, starting from rest. Determine its constant angular acceleration and the time required?

Answer 1

solution;

$final angular velocity w_{f}=30ras/s\\

initial w_{o}=0\\

w_{f}^{2}-w_{0}^{2}=2\alpha \theta \\

(30)^2=2.\alpha .20\times2\pi \\

\alpha =\frac{(30)^2}{2\times2\pi \times20}\\

\alpha =3.58rad/s^2\\

w_{f}=w_{o}+\alpha t\\

t=\frac{wf}{\alpha }=\frac{30}{3.58}\\

t=8.38$

Answer 2

Explanation:

It is given that,

Displacement, $\theta=2\ rev=12.56\ rad$

Initial angular velocity, $\omega_i=0$

Final angular velocity, $\omega_f=30\ rad/s$

Let $\alpha$ is its angular acceleration. Using the third equation of kinematics to find it as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\omega_f^2=2\alpha \theta$

$\alpha =\dfrac{\omega_f^2}{2\theta}$

$\alpha =\dfrac{(30)^2}{2\times 12.56}$

Angular acceleration, $\alpha =35.82\ rad/s^2$

Let t is the time required to find it as :

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

$t=\dfrac{30}{35.82}$

t = 0.83 seconds

So, the angular acceleration of the gear is $35.82\ rad/s^2$ and the time taken is 0.83 seconds.