Answer:
a) L = 2.10x10⁴⁰ kg*m²/s
b) τ = 1.12x10²⁴ N.m
Explanation:
a) The angular momentum (L) of the pulsar can be calculated using the following equation:
[tex] L = I \omega [/tex]
Where:
I: inertia momentum
ω: angular velocity
First we need to calculate ω and I. The angular velocity can be calculated as follows:
[tex] \omega = \frac{2 \pi}{T} [/tex]
Where:
T: is the period = 33.5x10⁻³ s
[tex] \omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s [/tex]
The inertia moment of the pulsar can be calculated using the following relation:
[tex] I = \frac{2}{5}mr^{2} [/tex]
Where:
m: is the mass of the pulsar = 2.8x10³⁰ kg
r: is the radius = 10.0 km
[tex] I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2} [/tex]
Now, the angular momentum of the pulsar is:
[tex] L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1} [/tex]
b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:
[tex] \tau = I*\alpha [/tex]
Where:
α: is the angular acceleration = 10⁻¹⁴ rad/s²
[tex]\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m[/tex]
I hope it helps you!
