Question

A barrel ride at an amusement park starts from rest and speeds up to 0.520 rev/sec in 7.26 s. What is the angular acceleration during that time? (Unit = rad/s^2)

Answer 1

Answer:

Angular acceleration of the barrel is 0.011 rad/s².

Explanation:

It is given that,

Initial speed of the barrel ride = 0

Final speed of the barrel ride, $\omega=0.52\ rev/sec$

On converting rev/sec to rad/sec as :

Since, 1 revolution = 2π radian

So, $\omega=0.52\ rev/sec=0.082\ rad/sec$

Time, t = 7.26 s

We need to find the angular acceleration of the barrel during that time. It is given by :

$\alpha=\dfrac{d\omega}{dt}$

$\alpha=\dfrac{0.082\ rad/s}{7.26\ s}$

$\alpha=0.011\ rad/s^2$

So, the angular acceleration of the barrel is 0.011 rad/s². Hence, this is the required solution.