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, [tex]\omega=0.52\ rev/sec[/tex]
On converting rev/sec to rad/sec as :
Since, 1 revolution = 2π radian
So, [tex]\omega=0.52\ rev/sec=0.082\ rad/sec[/tex]
Time, t = 7.26 s
We need to find the angular acceleration of the barrel during that time. It is given by :
[tex]\alpha=\dfrac{d\omega}{dt}[/tex]
[tex]\alpha=\dfrac{0.082\ rad/s}{7.26\ s}[/tex]
[tex]\alpha=0.011\ rad/s^2[/tex]
So, the angular acceleration of the barrel is 0.011 rad/s². Hence, this is the required solution.
