Respuesta :
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
[tex]\alpha=\frac{\Delta \omega}{\Delta t}[/tex]
where
[tex]\Delta \omega[/tex] is the change in angular velocity
[tex]\Delta t[/tex] is the time elapsed
The angular velocity can be written as
[tex]\omega=\frac{2\pi}{T}[/tex]
where T is the period of rotation of the object.
Therefore, the change in angular velocity can be written as
[tex]\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})[/tex]
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
[tex]T'=T+8.09\cdot 10^{-6} =0.01400809 s[/tex]
Therefore, the change in angular velocity after 1 year is
[tex]\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s[/tex]
So, the angular acceleration of the pulsar is
[tex]\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y[/tex]
b)
To solve this part, we can use the following equation of motion:
[tex]\omega'=\omega + \alpha t[/tex]
where
[tex]\omega'[/tex] is the final angular velocity
[tex]\omega[/tex] is the initial angular velocity
[tex]\alpha[/tex] is the angular acceleration
t is the time
For the pulsar in this problem:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s[/tex] is the initial angular velocity
[tex]\omega'=0[/tex], since we want to find the time t after which the pulsar stops rotating
[tex]\alpha = -0.259 rad/s/y[/tex] is the angular acceleration
Therefore solving for t, we find the time after which the pulsar stops rotating:
[tex]t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y[/tex]
c)
As we said in the previous part of the problem, the rate of change of the period of the pulsar is
[tex]\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y[/tex]
which means that the period of the pulsar increases by
[tex]\Delta T=8.09\cdot 10^{-6} s[/tex]
For every year:
[tex]\Delta t=1 y[/tex]
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
[tex]T=T_0+\frac{\Delta T}{\Delta t}\Delta t[/tex]
where [tex]T_0[/tex] is the original period and
[tex]\Delta t=869 y[/tex]
is the time that has passed; solving for T0,
[tex]T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s[/tex]
The pulsar's angular acceleration is [tex]\mathbf{ -8.21 \times 10^{-9} \ rad/s^2}[/tex]. 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;
[tex]\mathbf{\omega = \dfrac{2 \pi }{T}}[/tex]
The angular acceleration is the changes in the angular speed with time and it can be expressed by the formula:
[tex]\mathbf{\alpha = \dfrac{\omega _2 -\omega _1}{t}}[/tex]
a.
From the given information;
The angular speed for the pulsar at present [tex]\mathbf{\omega _1 = \dfrac{2 \pi}{T_1}}[/tex]
[tex]\mathbf{\omega _1 = \dfrac{2 \times 3.14}{0.0140 \ s}}[/tex]
[tex]\mathbf{\omega _1 = 448.5714286 \ rad/s }[/tex]
The angular speed for the pulsar after 1 year [tex]\mathbf{\omega _2 = \dfrac{2 \pi}{T_2} }[/tex]
[tex]\mathbf{\omega _2 = \dfrac{2 \times 3.14}{(0.0140 +8.09 \times 10^{-6} )} }[/tex]
[tex]\mathbf{\omega _2 =448.3123681 \ rad/s }[/tex]
The angular acceleration [tex]\mathbf{\alpha = \dfrac{\omega _2 -\omega_1}{1 \ year }}[/tex]
[tex]\mathbf{\alpha = \dfrac{448.3123681 \ rad/s -448.5714286 \ rad/s}{3.156*10^7 \ s}}[/tex]
[tex]\mathbf{\alpha = -8.21 \times 10^{-9} \ rad/s^2}[/tex]
The negative sign of the angular acceleration shows that the motion of the pulsar is in a backward motion.
b.
The first kinematics of angular motion can be expressed as:
[tex]\mathbf{\omega = \omega _1 + \alpha t}[/tex]
[tex]\mathbf{t= \dfrac{\omega - \omega_1}{\alpha}}[/tex]
Given that the final angular speed is zero [tex]\mathbf {\omega = 0}[/tex], then:
[tex]\mathbf{t= \dfrac{0 - 448.5714286 \ rad/s}{-8.21 \times 10^-9}}[/tex]
t = 5.46 × 10¹⁰ s
t = 1726.87 years
c.
Suppose the initial angular speed be [tex]\mathbf{\omega _o}[/tex] and the time [tex]t_o[/tex] = 869 years ago.
Then,
[tex]\mathbf{\omega _1 = \omega _o+ \alpha t_o}[/tex]
[tex]\mathbf{ \omega _o= 448.5714286 \ rad/s - (-8.21 \times 10^{-9} \ rad/s) \times 869 \ years \times 3.156 \times 10^7}[/tex]
[tex]\mathbf{ \omega _o= 673.74 \ rad/s}[/tex]
Now, the initial time period [tex]\mathbf{T_o = \dfrac{2 \pi}{\omega _o}}[/tex]
[tex]\mathbf{T_o = \dfrac{2 \pi}{673.74 \ rad/s}}[/tex]
[tex]\mathbf{T_o = 0.0093 \ s}[/tex]
Learn more about angular speed and acceleration here:
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