A certain star of radius 6.34 million meters, rotates twice around its axis in o
ID: 1487948 • Letter: A
Question
A certain star of radius 6.34 million meters, rotates twice around its axis in one earth day. Find its angular frequency. This star has a mass of 1.00x1032 kg. When this star collapses to a neutron star, its radius shrinks by a factor of 104 while it loses 30% of its mass. Find the angular velocity of this star after the collapse. Assume that the spherical shape and density of the star remain unchanged . Find the speed and centripetal acceleration of a particle on the equator of the newly formed neutron star.
Answer: =2.07E4/s v=1.31E7 m/s a=2.72E11 m/s2
Explanation / Answer
given data
radius of star, r = 6.34*10^6 m
M = 1*10^32 kg
time period, T = 0.5 day = 0.5*24*60*60 = 43200 s
a) initila angular speed, w1 = 2*pi/T
= 2*pi/43200
= 1.454*10^-4 rad/s
Initial moment of Inetia, I1 = (2/5)*M*r^2
when 30% of mass is lost,
remainig mass = 0.7*M
radius = r/10^4
final moment of inertia, I2 = (2/5)*0.7*M*(r/10^4)^2
= (2/5)*M*r^2*(0.3*10^-8)
let w2 is the final angular speed.
Apply, I1*w1 = I2*w2
==> w2 = I1*w1/I2
= (2/5)*M*r^2*1.454*10^-4/((2/5)*M*r^2*(0.7*10^-8))
= 1.454*10^-4/(0.7*10^-8)
= 2.077*10^4 rad/s <<<<<<<<<---------------Answer
B) linear speed of a point on the diamter, v2 = r2*w2
= (6.34*10^6/10^4)*2.077*10^4
= 1.317*10^7 m /s <<<<<<<<<---------------Answer
C) a_rad = v2^2/r2
= (1.317*10^7)^2/(6.34*10^6/10^4)
= 2.73*10^11 m/s^2 <<<<<<<<<---------------Answer
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