When a large star becomes a supernova, its core may be compressed so tightly tha

Question

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 37 km (about the size of the San Francisco area). If a neutron star rotates once every second,a) what is the speed of a particle on the star's equator?

b)What is the particle's centripetal acceleration in meters per second squared?

c)What is the particle's centripetal acceleration in g-units (ratio to g)?

d)If the neutron star rotates even faster, what happens, i.e. what would be the particle's centripetal acceleration in g-units if the period was 0.5 s?explanation if possible

Explanation / Answer

This question is really just to make sure you're comfortable with changing angular velocity to tangental velocity.
a) Angular velocity, omega, can be rewritten as (2*pi/T), where pi is 3.14 and T is the period (1 second)
Omega= 2 (s^-1)
Next, we need to relate omega to velocity. By multiplying omega by the radius, (37000 meters), we can get tangental velocity. This identity is derived from changing an angle into an arc length.

So from this, we get that the speed is 74,000*pi m/s
That's really, really fast.

b) To get centripetal acceleration, we can use another identity that comes out of a really nifty pro
The proof tells us that acentripetal=v^2/r
Therefore centripetal acceleration = 1.46 × 10^6 m^2/s

c) To put this into terms of g's, just divide by the acceleration due to gravity. g=9.81
Centripetal acceleration = 1.49×10^5 m^2/s

d) T = 0.5 s
Omega = 4 (s^-1)
Velocity = 14800 m/s
Centripetal acceleration = v^2/r = 5.48×10^6 m^2/s
Centripetal acceleration in g units = 5.96 ×10^5 m^2/s

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