Rotational Motion (Neutron Star)

AI Thread Summary
The discussion focuses on a physics homework problem involving rotational motion of a star and its transformation into a neutron star. The calculations for angular speed, angular momentum, and the radius of the neutron star are presented, but there's a suggestion to revise the angular velocity calculation to use radians instead of degrees. Additionally, clarification is provided that the torque calculation in part (iv) should pertain to the neutron star's current state rather than its initial form. Overall, the responses emphasize the importance of correct unit usage and the context of the calculations.
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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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