Physics: 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?

To solve this, you would divide 5 from 1, to get 1/5 cups of sugar...

A neutron star is an extremely dense, rapidly spinning

gonzalesalexiaouv1bg

20.01.2022

(a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Explanation:

Given that,

Mass of neutron $M_{n}= 13M_{s}$

Density of neutron $\rho=4.8\times10^{17}\ kg/m^3$

(a). We need to calculate the rotational inertia

Using formula of rotational inertia for sphere

$I=\dfrac{2}{5}MR^2$ ...(I)

We know that,

$\rho=\dfrac{M}{V}$

Put the value of volume

$\rho=\dfrac{3M_{n}}{4\pi R^3}$

$R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value of R in equation (I)

$I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value into the formula

$I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}$

$I=5.72\times10^{39}\ kg m^2$

The rotational inertia is $5.72\times10^{39}\ kg m^2$ .

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

Using formula of torque

$\tau=I\times \alpha$

Put the value into the formula

$\tau=5.72\times10^{39}\times5.6\times10^{-5}$

$\tau=3.20\times10^{35}\ N-m$

The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Hence, (a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

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