Question

a tire rolls from rest to an angular velocity of 6 radians per second. the angle it turns through is 12 radians. what is the angular acceleration of the tire?

Answer 1

Answer:

$\alpha =1.5\ rad/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega_i=0$

Final angular speed, $\omega_f=6\ rad/s$

Angular displacement, $\theta=12\ rad$

We need to find the angular acceleration of the tire. We can find it using the third equation of rotational kinematics. So,

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{6^2-0^2}{2\times 12}\\\\\alpha =1.5\ rad/s^2$

So, the angular acceleration of the tire is equal to $1.5\ rad/s^2$.