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Iteru [2.4K]
3 years ago
5

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 = 0.0140 s that is increasing at the rate of 8.09 x 10-6 s/y. (a) What is the pulsar's angular acceleration ? (b) If is constant, how many years from now will the pulsar stop rotating? (c) Suppose the pulsar originated in a supernova explosion seen 869 years ago. Assuming constant , find the initial T.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

\alpha=\frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

The angular velocity can be written as

\omega=\frac{2\pi}{T}

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

T'=T+8.09\cdot 10^{-6} =0.01400809 s

Therefore, the change in angular velocity after 1 year is

\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s

So, the angular acceleration of the pulsar is

\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y

b)

To solve this part, we can use the following equation of motion:

\omega'=\omega + \alpha t

where

\omega' is the final angular velocity

\omega is the initial angular velocity

\alpha is the angular acceleration

t is the time

For the pulsar in this problem:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s is the initial angular velocity

\omega'=0, since we want to find the time t after which the pulsar stops rotating

\alpha = -0.259 rad/s/y is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y

which means that the period of the pulsar increases by

\Delta T=8.09\cdot 10^{-6} s

For every year:

\Delta t=1 y

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

T=T_0+\frac{\Delta T}{\Delta t}\Delta t

where T_0 is the original period and

\Delta t=869 y

is the time that has passed; solving for T0,

T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s

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