A pulsar is a rapidly rotating neutron star that emits one radio pulse for each

Question

A pulsar is a rapidly rotating neutron star that emits one radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between two pulses.

At present, the pulsar in the central region of the Crab nebula has a rotation period of T = 0.12000000 sec, and this is observed to be increasing at the rate of 0.00000538 sec per year.

What is the angular acceleration of the pulsar? If the angular acceleration of the pulsar is constant, how long will it take for the pulsar to stop rotating?

Explanation / Answer

Given  

a pulsar having

T = 0.12000000 sec so the initial angular velocity is Wi = 2pi /T = 2pi /(0.12000000) rad/s = 52.36 rad/s

observed to be increasing at the rate of time 0.00000538 sec per year = dt

so after one year the time period of the pulsar is Tf = Ti+dt = 0.12+0.00000538 s = 0.12000538 s

we know that the angular acceleration is alpha = 2pi /(Tf-Ti)

alpha = 2pi (1/Tf -1/Ti)

and angular acceleration per year is  

we know that one year time = 365*24*60*60 s = 31536000 s

so theangular acceleration is alpha = 2pi (1/Tf -1/Ti) / 31536000

alpha = 2pi(1/0.12000538 - 1/0.12) /(31536000 ) rad/s2

alpha = (-7.4434379878)*10^-11 rad/s2

now the time taken for the pulsar to stop rotating is from equation Wf = wi - alpha*t

here Wf = 0 , wi = 52.36 rad/s ,

t = -wi/alpha

t = -(52.36)/((-7.4434379878)*10^-11) s

t= 703438385405 s

t = 22305.89 years

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