Question

Please solve for Tf and how fast the neuron star rotate in rotations per seconds.

Formation of a Neutron Star A star rotates with a period of 29 days about an axis through its center. The period is the time interval required for a point on the star's equator to make one complete revolution around the axis of rotation. After the star undergoes a supernova explosion, the stellar core, which had a radlus of 1.5 x 104 km, collapses into a neutron star of radius 4.9 km. Determine the perlad of rotation of the neutron star SOLUTION Conceptualize The change in the neutron star's motion is similar to that of the skater described carlier in the textbook, but in the reverse direction. As the mass of the star moves closer to the rotation axis, we expect the star to spin faster Categorize Let us assume that during the collapse of the stellar core, (1) no external torque acts on it, (2) it remains spherical with the same relative mass distribution, and (3) its mass remains constant. We categorize the star as an isolatedsystem in terms of angular momentum. we do not know the mass distribution of the star, but we have assumed the distribution is symmetric, so the moment of inertia can be expressed as kMR2, where k is some numerical constant. (From this table, for example, we see that k-_ for a solid sphere and k-_ for a soherical shell.) Analyze (Use the following as necessary: ω, T., Tp Rjr RP k, and M.) 2m Let's use the symbol T for the period, with T, being the initial period of the star and Tf being the period of the neutron star. The star's angular speed is given by co- From the Isolated system model for angular momentum, write the following equation for the star: Use 2T to rewrite this equation in terms of the initial and final periods: '주-愕) Substitute the moments of inertia in the preceding cquation 2x Solve for the final period of the star: Substitute numerical values to find the final period (In s) Tf267.4 Finalize The neutron star does indeed rotate taster after it collapses, as predicted. It moves very fast, in fact, rotating (in rotatons per second) times each second.

Answer 1

Initial radius , Ri = 1.5e4 Km

Final radius, Rf = 4.9 Km

Initial time period = 29 days = 29 * 24 hours * 60 minutes * 60 seconds = 2505600 sec

so,

Tf = (Rf / Ri)² * Ti

Tf =  (4.9 / 15000 )² * 2505600

Tf = 0.267 seconds

Rotations per second will be the frequency

which is around 3.745 rotations / sec