Rotational Motion (Neutron Star)

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SUMMARY

The discussion focuses on the calculations related to the rotational motion of a neutron star derived from a massive star with mass Mi = 3.0 × 1030 kg and radius Ri = 7.0 × 108 m. The angular speed of the star is calculated as ω = 2.46 × 10-5 rad/s, and its angular momentum is L = 1.446 × 1043 kgm2/s. Upon collapsing into a neutron star, the final radius is determined to be approximately 34.71 km, with a torque of τ = 1 × 1039 N acting on the star due to its observed slowing period of rotation.

PREREQUISITES
  • Understanding of angular velocity and angular momentum
  • Familiarity with the conservation of angular momentum principle
  • Knowledge of rotational dynamics equations
  • Basic proficiency in unit conversions (e.g., seconds to years)
NEXT STEPS
  • Study the derivation of angular momentum in rotating systems
  • Learn about the properties and formation of neutron stars
  • Explore the effects of torque on rotating bodies
  • Investigate the relationship between period of rotation and angular speed
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Students preparing for physics exams, astrophysicists studying stellar evolution, and educators teaching concepts of rotational motion and dynamics.

patrickmoloney
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Homework Statement


I'm doing a question from a past paper, preparing for an upcoming exam. There is no solutions so I was wondering if my answer is correct for all parts.

Take a star to be a uniform sphere with mass M_{i}=3.0 \times 10^{30} Kg and radius R_{i} = 7.0 \times 10^{8}m that rotates with a period of 27.0 days.

(i) What's the star's angular speed of rotation?
(ii) What's the star's angular momentum?
(iii) At the end of it's lifetime, the star collapses to form a very compact star called a neutron star. If the star retains all it's mass when it collapses and the angular speed of the rotation of the neutron star that forms is \omega _{f} = 10^{4} rad/s. what is the radius of the neutron star in kilometres?
(iv) If the neutron star's period of rotation is observed to be increasing at a rate of 1.2 \times 10^{-5}s/yr, what is the torque acting on the star?

Homework Equations

The Attempt at a Solution


(i) 27 days = 2332800 s. The star rotates 360 degrees in 27 days. \omega = \frac{\frac{360}{2\pi}}{T} = 2.46 \times 10^{-5} rad/s

(ii) \begin{align} L &amp; =I \omega \\<br /> &amp; = \frac{2}{5} M_{i}R_{i}^{2} \omega \\<br /> &amp; = \frac{2}{5}(3 \times 10^{30})(7 \times 10^{8})^{2} \\<br /> &amp; = 1.446 \times 10^{43} Kgm^{2}/s\end{align}

(iii) Conservation of angular momentum \begin{align} I_{i} \omega_{i}&amp;= I_{f} \omega_{f} \\<br /> \frac{2}{5}M_{i}R_{i}^{2} \omega_{i} &amp; = \frac{2}{5}M_{f}R_{f}^{2} \omega_{f} \\<br /> R_{f} &amp; = \sqrt{\frac{\omega_{i}}{\omega_{f}}R_{i}^{2}} \\<br /> &amp; = 34.718.87 m \\ <br /> &amp; = 34.71 km \end{align}

(iv) \begin{align} \Delta L &amp; = I_{f} \omega_{f} - I_{i} \omega{i} \\<br /> &amp; = 1.2 \times 10^{34} kgm^{2}/s\end{align}

\tau = \frac{\Delta L}{\Delta T} = \frac{1.2 \times 10^{34}}{1.2 \times 10^{-5}} = 1 \times 10^{39} N

Thanks. Does this look okay?
 
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Revisit your determination of the angular velocity in part (i). The method should not involve degrees at all. One "cycle" or rotation is ##2 \pi ~ rad## (just as one rotation is also 360 degrees, but your angular velocity should be specified in radians per second).

I believe that part (iv) is meant to pertain to the star once it's already in neutron star form: it's observed to be slowing at the specified rate. So it won't involve the initial state of the star at all.
 

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