Physics, 24.03.2020 21:58 mimireds5419

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.

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The pulsar's angular acceleration is $\mathbf{ -8.21 \times 10^{-9} \ rad/s^2}$ . If it's constant, the pulsar will stop rotating after 1726.87 years. The initial time period T after the explosion is seen 869 years ago will be = 0.0093 s

The angular speed tells us about the angle at which an object moves through space within a specified range of time.

The formula used for calculating the angular speed can be expressed as;

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

The angular acceleration is the changes in the angular speed with time and it can be expressed by the formula:

$\mathbf{\alpha = \dfrac{\omega _2 -\omega _1}{t}}$

From the given information;

The angular speed for the pulsar at present $\mathbf{\omega _1 = \dfrac{2 \pi}{T_1}}$

$\mathbf{\omega _1 = \dfrac{2 \times 3.14}{0.0140 \ s}}$

$\mathbf{\omega _1 = 448.5714286 \ rad/s }$

The angular speed for the pulsar after 1 year $\mathbf{\omega _2 = \dfrac{2 \pi}{T_2} }$

$\mathbf{\omega _2 = \dfrac{2 \times 3.14}{(0.0140 +8.09 \times 10^{-6} )} }$

$\mathbf{\omega _2 =448.3123681 \ rad/s }$

The angular acceleration $\mathbf{\alpha = \dfrac{\omega _2 -\omega_1}{1 \ year }}$

$\mathbf{\alpha = \dfrac{448.3123681 \ rad/s -448.5714286 \ rad/s}{3.156*10^7 \ s}}$

$\mathbf{\alpha = -8.21 \times 10^{-9} \ rad/s^2}$

The negative sign of the angular acceleration shows that the motion of the pulsar is in a backward motion.

The first kinematics of angular motion can be expressed as:

$\mathbf{\omega = \omega _1 + \alpha t}$

$\mathbf{t= \dfrac{\omega - \omega_1}{\alpha}}$

Given that the final angular speed is zero $\mathbf {\omega = 0}$ , then:

$\mathbf{t= \dfrac{0 - 448.5714286 \ rad/s}{-8.21 \times 10^-9}}$

t = 5.46 × 10¹⁰ s

t = 1726.87 years

Suppose the initial angular speed be $\mathbf{\omega _o}$ and the time t_o = 869 years ago.

Then,

$\mathbf{\omega _1 = \omega _o+ \alpha t_o}$

$\mathbf{ \omega _o= 448.5714286 \ rad/s - (-8.21 \times 10^{-9} \ rad/s) \times 869 \ years \times 3.156 \times 10^7}$

$\mathbf{ \omega _o= 673.74 \ rad/s}$

Now, the initial time period $\mathbf{T_o = \dfrac{2 \pi}{\omega _o}}$

$\mathbf{T_o = \dfrac{2 \pi}{673.74 \ rad/s}}$

$\mathbf{T_o = 0.0093 \ s}$

Learn more about angular speed and acceleration here:

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Answer from: chell6170

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

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

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

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:

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

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,

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

Answer from: Quest

what is the question because that is not a question.

Answer from: Quest

answer: b.

explanation:

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