Answer:
0.0768 revolutions per day
Explanation:
R = Radius
[tex]\omega[/tex] = Angular velocity
As the mass is conserved the angular momentum is conserved
[tex]I_1\omega_1=I_2\omega_2\\\Rightarrow \frac{I_1}{I_2}=\frac{\omega_2}{\omega_1}[/tex]
Moment of intertia for solid sphere
[tex]I_1=\frac{2}{5}MR^2\\\Rightarrow I_1=0.4MR^2[/tex]
Moment of intertia for hollow sphere
[tex]I_2=\frac{2}{3}M(4.3R)^2\\\Rightarrow I_2=12.327MR^2[/tex]
Dividing the moment of inertia
[tex]\frac{I_1}{I_1}=\frac{0.4MR^2}{12.327MR^2}\\\Rightarrow \frac{I_1}{I_2}=0.032[/tex]
From the first equation
[tex]\omega_2=\omega_1\frac{I_1}{I_2}\\\Rightarrow \omega_2=2.4\times 0.032\\\Rightarrow \omega_2=0.0768\ rev\day[/tex]
The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day