Question

A star is rotating at 5.55x10-5 π rad/s, its velocity decelerates at a constant rate of 0.5x10-9 π rad/s2.?

Answer 1

The current rotational period of that star is 10.01 hours.

Explanation:

Given that,

Initial angular velocity of the star, $\omega=5.55\times 10^{-5}\pi \ rad/s$

It decelerates, final angular speed, $\omega_f=0$

Deceleration, $\alpha =-0.5\times 10^{-9}\pi \ rad/s^2$

It is not required to use the rotational kinematics formula. The angular velocity in terms of time period is given by :

$\omega=\dfrac{2\pi}{T}$

T is current rotational period of that star

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{5.55\times 10^{-5}\pi \ rad/s}$

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.