Question

A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a massive star at the end ofits life. A neutron star with 13 times the Sun's mass has an essentially uniform density of 4.8 x 1017 kg/ m3. (a) What's itsrotational inertia? (b) The neutron star's spin rate slowly de-creases as a result of torque associated with magnetic forces. Ifthe spin-down rate is 5.6 x 10-5 rad/s2, what's the magnitude of the magnetic torque?

Answer 1

(a). The rotational inertia is
`5.72*10^(39)\ kg m^2`

(b). The magnitude of the magnetic torque is
`3.20*10^(35)\ N-m`

Step-by-step explanation:

Given that,

Mass of neutron
`M_(n)= 13M_(s)`

Density of neutron
`\rho=4.8*10^(17)\ kg/m^3`

(a). We need to calculate the rotational inertia

Using formula of rotational inertia for sphere

`I=(2)/(5)MR^2`...(I)

We know that,

`\rho=(M)/(V)`

Put the value of volume

`\rho=(3M_(n))/(4\pi R^3)`

`R^2=((3M_(n))/(4\pi\rho))^{(2)/(3)}`

Put the value of R in equation (I)

`I=(2)/(5)* M_(n)*((3M_(n))/(4\pi\rho))^{(2)/(3)}`

Put the value into the formula

`I=(2)/(5)*(13*2*10^(30))^{(5)/(3)}*((3)/(4\pi*(4.8*10^(17))))^{(2)/(3)}`

`I=5.72*10^(39)\ kg m^2`

The rotational inertia is
`5.72*10^(39)\ kg m^2`.

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

`\tau=I* \alpha`

Put the value into the formula

`\tau=5.72*10^(39)*5.6*10^(-5)`

`\tau=3.20*10^(35)\ N-m`

The magnitude of the magnetic torque is
`3.20*10^(35)\ N-m`

Hence, (a). The rotational inertia is
`5.72*10^(39)\ kg m^2`

(b). The magnitude of the magnetic torque is
`3.20*10^(35)\ N-m`