Question

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. Suppose a pulsar has a period of rotation of T = 0.0922 s that is increasing at the rate of 3.64 x 10-7 s/y. (a) What is the pulsar's angular acceleration

Answer 1

Answer:

$-8.54883\times 10^{-12}\ rad/s^2$

Explanation:

T = Time period = 0.0922 s

$\dfrac{dT}{dt}=3.64\times 10^{-7}\ s/y$

Angular speed is given by

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

Angular acceleration is given by

$\alpha=\dfrac{d\omega}{dt}\\\Rightarrow \alpha=\dfrac{d\dfrac{2\pi}{T}}{dt}\\\Rightarrow \alpha=-\dfrac{2\pi}{T^2}\dfrac{dT}{dt}\\\Rightarrow \alpha=-\dfrac{2\pi}{0.0922^2}\times\dfrac{3.64\times 10^{-7}}{365.25\times 24\times 60\times 60}\\\Rightarrow \alpha=-8.52541\times 10^{-12}\ rad/s^2$

The pulsar's angular acceleration is $-8.52541\times 10^{-12}\ rad/s^2$

Answer 2

Answer:

- 8.5 x 10^-12 rad/s²

Explanation:

Time period, T = 0.0922 s

dT / dt = 3.64 x 10^-7 s/year

As we know that

α = dω/dt

where, α is the angular acceleration and ω is the angular velocity

ω = 2π/T

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

$\alpha =\frac{d\left ( \frac{2\pi }{T} \right ) }{dt}$

$\alpha =-\frac{2\pi}{T^{2}}\times \frac{dT}{dt}$

By substituting the values

$\alpha =-\frac{2\times 3.14}{0.0922^{2}}\times \frac{3.64\times 10^{-7}}{365\times 24\times 3600}$

α = - 8.5 x 10^-12 rad/s²