Centripetal Acceleration of pulsar

[LeakyFrog]

Homework Statement

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from pulsars at regular intervals equal to the period that they rotate. A certain pulsar has a period currently of length 33.085m/s and is estimated to have an equatorial radius of 15km.
a) What is the value of the centripetal acceleration of an object on the surface at the equator of the pulsar?
b) many pulsars are observed to have periods that lengthen slightly with time. The rate of slowing of this pulsar is 3.5x10^-13 seconds, which implies that if this rate remains constant it will stop spinning in 9.5x10^10 seconds. What is the tangential acceleration of an object on the equator of this neutron star?

Homework Equations

a_c = v²/r
a_t= dv/dt
v=2(pi)r/T

The Attempt at a Solution

a) for a) I got 5.413x10^8 (m/s²)
b) This is the part that I'm getting stuck on. I'm not really sure what tangential acceleration is. And even using the dv/dt I'm not entirely sure where I would get dv/dt. Any suggestions/questions for this portion are appreciated

 

[rock.freak667]

the rate of slowing means the rate of change of angular displacement.

So you can find the angular acceleration and then use a_t=αr

 

[tomcue]

.

I would like to first clarify that the given equation for centripetal acceleration (ac = v^2/r) is only applicable for objects moving in circular motion. In the case of a pulsar, which is a rapidly rotating neutron star, the motion is not necessarily circular and the concept of centripetal acceleration may not be entirely applicable.

For part a), assuming that the given period of 33.085m/s refers to the time taken for one full rotation, we can calculate the tangential velocity of an object on the equator using the equation v=2(pi)r/T. This gives us a value of 2.97x10^8 m/s. Now, using the equation ac = v^2/r, we can calculate the centripetal acceleration at the equator of the pulsar to be 5.413x10^8 m/s^2.

For part b), the given rate of slowing of the pulsar (3.5x10^-13 seconds) is the change in the period with respect to time. This can be written as dv/dt, where v is the tangential velocity. Using the same equation as before, we can calculate the initial tangential velocity of the pulsar to be 2.97x10^8 m/s. Now, using the given time of 9.5x10^10 seconds, we can calculate the change in velocity (dv) to be 3.33x10^-5 m/s. Finally, using the equation at=dv/dt, we can calculate the tangential acceleration at the equator of the pulsar to be 3.5x10^-18 m/s^2.

It is important to note that the concept of tangential acceleration may not be entirely applicable in the case of a pulsar, as the motion is not necessarily circular and the pulsar is also undergoing other complex changes and processes. Further research and study may be needed to fully understand the dynamics of a pulsar's motion.