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 = 0.0786 s that is increasing at the rate of 7.03 x 10-6 s/y. (a) What is the pulsar's angular acceleration α

Answer 1

Answer:

$\alpha =-2.2669642\times^{-10}rad/s^2$

Explanation:

Angular acceleration is defined by $\alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}$

Angular velocity is related to the period by $\omega=\frac{2\pi}{T}$

Putting all together:

$\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})$

Taking our initial (i) point now and our final (f) point one year later, we would have:

$\Delta t=1\ year=(365)(24)(60)(60)s=31536000 s$

$T_i=0.0786s$

$T_f=0.0786s+7.03\times10^{-6}s$

So for our values we have:

$\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2$

Where the minus sign indicates it is decelerating.