question archive A star with an initial radius of 1
A star with an initial radius of 1
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A star with an initial radius of 1.00 10^8 m and period of 30.0 days collapses suddenly to a radius of 1.00 10^4 m.
(a) Find the period of rotation after collapse.
(b) Find the work done by gravity during the collapse if the mass of the star is 2.00 1030 kg.(The only force doing work on the system is gravitation, so the work done by gravitation is the change in kinetic energy of the star. )
(c) What is the speed of an indestructible person standing on the equator of the collapsed star? (Neglect any relativistic or thermal effects, and assume the star is spherical before and after it collapses).
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Answer:
a)moment of inertia of star I = (2/5) MR2
from this ,
I1/I2 = [(2/5) MR12 ]/[(2/5) MR22]
= (R1/R2)2
= (1*108 m / 1*104 m)2
= 1*108
(or) I2 = (1*10-8)I1
from law of conservation of angular momentum ,
I1ω1 = I2ω2
ω2 = (I1/I2)ω1
= (1*108)ω1
ω2/ω1 = 1*108
(2π/T2) / (2π/T1) = 1*108
time period T2 = (1*10-8)T1
= (1*10-8)(30 days)
= (1*10-8)(30*24*60*60) s
˜ 2.6*10-2 s
b)
work done by gravity = change in kinetic energy
ΔKE = 1/2 I2 ω22 -1/2 I1 ω12
= (1/2) I2 ω2ω2 -(1/2) I1 ω1ω1 (since, I1ω1 = I2ω2 )
= (1/2) [I2 ω2]ω2 -(1/2) [I1 ω1]ω1
= (1/2) [I1 ω1](1*108)ω1 - (1/2) [I1 ω1]ω1
= (1/2) [I1 ω1] [(1*108)ω1 - ω1 ] ........................ (1)
here , ω1 = 2π/T1 = 2π/(30*24*60*60 s) = 2.42*10-6 rad/s
I1 = (2/5)MR12 = (2/5)(2*1030 kg)(1*108 m)2 = 0.8*1046 kg.m2
substitute the given data in eq (1) , we get
ΔKE = (1/2) [(0.8*1046)(2.42*10-6)] [(1*108)(2.42*10-6) -(2.42*10-6 )]
= (1/2) [(0.8*1046)(2.42*10-6)] [242 - 0.00000242]
= (1/2) [(0.8*1046)(2.42*10-6)] [241.99]
= 234.25*1040 J
(OR)
ΔKE = 2.34*1042 J
c)
speed V2 = R2ω2
= (1*104 m)((1*108)ω1)
= (1*104 m)((1*108)( 2.42*10-6 rad/s))
= 2.42*106 m/s
