Question

the rotating rod starts from rest and acquires a rotational speed n = 600 /minute in 2 second
with constant angular acceleration. The angular acceleration of the rod ​

Answer 1

Answer:

angular acceleration aplha = 10π rad/sec

Answer 2

Answer:

$\alpha=10\pi \ rad/s^2$

Explanation:

It is given that,

Initial speed of the rotating rod, $\omega_i=0$

Final speed of the rotating rod, $\omega_f = 600\times \dfrac{2\pi}{60}\ rev/min = 20\pi \ rad/s$

Time taken, t = 2 s

Let $\alpha$ is the angular acceleration of the rod. Using the first equation of rotational kinematics as :

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

$\alpha =\dfrac{20\pi-0}{2}$

$\alpha=10\pi \ rad/s^2$

So, the angular acceleration of the rod is $10\pi \ rad/s^2$. Hence, this is the required solution.