## Trending News

# A pulsar is a fast rotating star that emits a beam that we can observe and use to calculate the rotation perio

. A pulsar is a fast rotating star that emits a beam that we can observe and use to calculate the rotation period of the pulsar. The Crab nebula pulsar has a period of 0.033 sec and is increasing at a rate of 1.26 x 10-5 sec/year. (a) What is the pulsar’s angular acceleration? (b) Assuming a constant angular acceleration, when the pulsar stop spinning? (c) The pulsar originated in a supernova explosion observed in the year 1054. What was the initial period of the pulsar?

### 2 Answers

- PhysicsDudeLv 71 decade agoFavorite Answer
Pulsar's period, T = 0.033 sec and is increasing at a rate of 1.26 x 10^-5 sec per year

(a) Angular acceleration, α = ∆ω/∆t

where ω = 2π/T = 6.283/0.033 = 190.39 rad/sec

Since the period increases by 1.26 x 10^-5 sec per year, then

∆T = 1.26 x 10^-5 sec => T' = T + ∆T = 0.0330126 sec

ω' = 2π/T' = 6.283/0.0330126 = 190.32 rad/sec

Hence = ∆ω = ω' - ω = -0.07267

Now, α = ∆ω/∆t = -0.07267/31536000 = 2.3 x 10^-9 rad/sec^2

(b) ω' = ω + αt, set ω' = 0, solve for t:

t = -ω/α = (190.39)/(2.3 x 10^-9) = 82,624,112,123.83 sec

or 2,620 years

(c) ω = ω(0) + αt, set t = (2006-1054)x365x24x60x60 sec

ω(0) = ω - αt = 190.39 - (-(2.3 x 10^-9)(30022272000))

ω(0) = 190.39 + 69.05 = 259.44 rad/sec or its initial period

T(0) = 0.024 sec

- Anonymous5 years ago
Hi, Guess instead of paying attention in class you were busy scratching your balls