Sunday

April 19, 2015

April 19, 2015

Posted by **shelby** on Monday, March 17, 2008 at 3:13pm.

- physics -
**Damon**, Monday, March 17, 2008 at 4:43pmW = 2 pi / T if W is angular velocity.

so W = 2 pi / .09 = 69.8 radians/second

Tin one year - T now = 4.67*10^- 6 seconds

but one year = 3600*24*365 = 3.15*10^7 seconds

so change of T per second = 4.67*10^-6 / 3.15*10^7

= 1.48 *10^-13 seconds/second = dT/dt

Now Either I do some calculus or you convert all that to angular velociteies by brute force.

I have dT/dt

I want dW/dt

but

W = 2 pi/T

dW/dT = 2 pi (-dT/dt)/T^2 =

2 pi (-1.48*10^-13) / (.09)^2

so angular accleration = dW/dt = a

a= - 1148 * 10^-13 radians /s^2

or

a= - 1.15 * 10^-10 radians/s^2

now you have the initial angular velocity and the angular acceleration and the number of seconds per year. I think you can take it from there

physics - A pulsar is a rapidly rotating neutron star that continuously emits a ...

Physics - A pulsar is a rapidly rotating neutron star that continuously emits a ...

Physics - The Crab Nebula One of the most studied objects in the night sky is ...

Physics - A pulsar is a rapidly rotating neutron star that continuously emits a ...

Physics (please help!!) - The Crab pulsar (m=2.00x10^30 kg) is a neutron star ...

Another astronomy (algebra-based physics) question - I’d prefer thorough ...

physics - Certain neutron stars (extremely dense stars) are believed to be ...

physics - Certain neutron stars (extremely dense stars) are believed to be ...

physics - A star with an initial radius of 1.0 x 10^8 m and period of 30.0 days ...

physics - A star with an initial radius of 1.0 x 10^8 m and period of 30.0 days ...