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.?

What is the current rotational period of that star (in hours)?

what is the formula i should use to solve this equation. Would it be the same as one of the rotational kinematics formula?

Answer 1

The current rotational period of that star is 10.01 hours.

Step-by-step explanation:

Given that,

Initial angular velocity of the star,
`\omega=5.55* 10^(-5)\pi \ rad/s`

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

Deceleration,
`\alpha =-0.5* 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=(2\pi)/(T)`

T is current rotational period of that star

`T=(2\pi)/(\omega)`

`T=(2\pi)/(5.55* 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.