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 18.0 km (about the size of a typical city). If a neutron star rotates once every 0.860 seconds, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

Answer 1

Answer:

Explanation:

a ) radius of star  R = 18 x 10³ m

time period of rotation T = .86 s.

angular velocity ω = 2π / T

= 2 X 3.14 / .86

= 7.3 radian / s .

velocity required

= ω R

7.3 x 18 x 10³

= 131.4 m /s

b )

centripetal acceleration

= ω² R

= ω R ω

= 131.4 x  7.3

959.22 m/s²

c )

If the neutron star rotates faster , both velocity of the particle on the equator

and centripetal acceleration increases.

Answer 2

a. The speed of the particle is 131.4 m /s

b. The magnitude of the particle's centripetal acceleration is 959.22 m/s²

c. It increased.

Calculation of the speed, magnitude:

a ) Since the radius of star  R = 18 x 10³ m

And,

time period of rotation T = .86 s.

So,

angular velocity ω = 2π / T

= 2 X 3.14 / .86

= 7.3 radian / s .

Now

velocity should be

= ω R

= 7.3 x 18 x 10³

= 131.4 m /s

b)

centripetal acceleration

= ω² R

= ω R ω

= 131.4 x  7.3

= 959.22 m/s²

c )

In the case when the neutron star rotates faster, both the velocity of the particle should be on the equator and  due to this, centripetal acceleration increases.

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