Pulsar Angular acceleration question

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SUMMARY

The discussion focuses on calculating the angular acceleration of a pulsar with a rotation period of 0.033 seconds and an increase in period of 1.26 x 10^(-5) seconds per year. The correct approach involves calculating the angular velocity using the formula ω = 2π/T, resulting in an initial angular velocity of 190.4 rad/s. The angular acceleration is determined by the change in angular velocity over time, requiring conversion from years to seconds for accurate results, leading to a negative angular acceleration value.

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Homework Statement


The problem is as follows. A certain pulsar (a rapidly rotating neutron star) has a period of rotation of 0.033s, and this is observed to be increased at a rate of 1.26(10^(-5)) seconds per year. what is the value of the angular acceleration in rad/s^2?


Homework Equations


w (omega) = 2pi/T. I calculated that to be 190.4.

angular acceleration = change in w / change in t.

The Attempt at a Solution



So, I assumed that the angular acceleration is constant. If this is so, then I should get my answer by calculating the angular velocity at T = 0.033s, and subtracting that calculated value with T= 0.033 + 1.26x10^(-5), and divide that number by 1.26x10^(-5). However, this didn't get me the right answer.

Any suggestions?
 
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The definition of angular acceleration is "rate of change of angular velocity over time."

So, I assumed that the angular acceleration is constant. If this is so, then I should get my answer by calculating the angular velocity at T = 0.033s, and subtracting that calculated value with T= 0.033 + 1.26x10^(-5)

True.

But you don't divide the result by 1.26(10^(-5)). You are going to end up with a angular acceleration in rad/year^2. You have to convert those year^2 to sec^2.

EDIT: You have to subtract T=0.033s to T= 0.033 + 1.26x10^(-5) s. So the angular acceleration is negative.


Source:

http://en.wikiversity.org/wiki/Torque_and_angular_acceleration

(at the very bottom)
 
Last edited:
Ahh yes ok, didn't catch that. Thanks!
 

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