Question

One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chinese in 1054. In 1968 it was discovered that a pulsar-a rapidly rotating neutron star that emits a pulse of radio waves with each revolution-lies near the center of the Crab nebula. The period of this pulsar is 33 ms. What is the angular speed in rad/s of the Crab nebula pulsar?

Answer 1

T = 33 ms = 0.033 s

ω = 2π/T

Answer 2

The angular speed of an object rotating with period T is 2*π/T; you have T = 33 m/s? m/s is a speed not a time. Do you mean 33 ms (milliseconds)? If so, ω = 2*π/0.033 = 190 rad/s

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further calculation ( not asked in question)

The average angular acceleration is ∆ω/∆t. ∆ω = ω1 - ω0

ω = 1/Tr where Tr is the time for one rotation. If Tr is the time for one rotation, then the time for one year of rotations T = 365*Tr

For 1906 T0 = 365*Tr0
for 2006 T1 = 365*Tr0 + 0.840

ω = 2*π/T

ω0 = 2*π/T0
ω1 = 2*π/T1

∆ω = 2*π/T1 - 2*π/T0 = 2*π/(365*Tr0) - 2*π/(365*Tr0 + 0.840)

Tr0 = 24*3600 sec

∆ω = 5.307*10^-15 rad/s

∆t = 100 yr = 100*365*24*3600 sec = 3.154*10^9 s

α = 5.307*10^-15/3.154*10^9 = 1.69*10^-24 rad/s²

In one rev the wheel travels one circumference = 2*π*r. At 0.373 revs/sec it travels 0.373 circumferences, v = 0.373*2*π*0.26 m/s = 0.609 m/s