Explanation:
Given that,
Initial angular velocity of the star, [tex]\omega=5.55\times 10^{-5}\pi \ rad/s[/tex]
It decelerates, final angular speed, [tex]\omega_f=0[/tex]
Deceleration, [tex]\alpha =-0.5\times 10^{-9}\pi \ rad/s^2[/tex]
It is not required to use the rotational kinematics formula. The angular velocity in terms of time period is given by :
[tex]\omega=\dfrac{2\pi}{T}[/tex]
T is current rotational period of that star
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\pi}{5.55\times 10^{-5}\pi \ rad/s}[/tex]
T = 36036.03 second
or
1 hour = 3600 seconds
So, T = 10.01 hours
So, the current rotational period of that star is 10.01 hours. Hence, this is the required solution. Hence, this is the required solution.
