C.O.M and Inertia of a sphere; top half twice as dense as lower half

[Kneemar]

Homework Statement

A sphere of radius L is made up of an upper hemisphere of uniform mass density σ = 2σ₀ and a lower hemisphere of uniform mass density σ = σ₀. Origin of co-ordinate system lies at centre of sphere with the denser hemisphere above the xy plane. Using spherical polar co-ordinates:

Homework Equations

(i) Find the mass of the sphere

(ii) Show the the centre of mass lies at z_COM = L/8

(iii) The moment of inertia for rotation of the sphere about the z-axis is defined as I_z = ∫∫∫l_z²dM, where l_z is the perpendicular distance of a mass element dM = σdV from the z-axis. Show that I_z = (4/5)(pi)σ₀L⁵

The Attempt at a Solution

(i) M = 4/3(pi)L³(1.5σ₀) = 2(pi)L³σ₀

(ii) z_COM = (∫zdM)/∫dM = (1/(2(pi)L³σ₀))*∫zdM

∫zdM = ∫zσdV
z = rcosθ
dV = r²sinθdrdθd$\varphi$
∫zdM = σ∫r³sinθcosθdpdθd$\varphi$

I could be doing okay up to this point, but not convinced.

Now what do I do considering I have the 2 different density hemispheres?

(iii) No attempt worth writing down



Cheers :)

ps. Sorry if my input style is horrible (particularly equations might be messy). It's my first post.

 

[mighty2000]

Hello!

Great job so far on your solution attempt! Let's take a look at each part and see if we can refine it a bit.

(i) Your approach is correct. You have correctly identified the mass of the sphere as the sum of the masses of the two hemispheres. However, your final answer should be M = (4/3)πL^3(2σ0 + σ0) = (4/3)πL^3(3σ0) = 4πL^3σ0.

(ii) You are on the right track with your approach. However, there are a few things to note. First, the integral for the x and y coordinates should also have a sinθ term in front, just like the z coordinate. This is because the x and y coordinates also vary with θ in spherical coordinates. Secondly, since the density is constant, it can be taken out of the integral. Finally, the limits of integration for r should be from 0 to L and for θ should be from 0 to π/2, since we are only considering the upper hemisphere. With these corrections, you should get zCOM = (3L/8).

(iii) For this part, we can use the parallel axis theorem to simplify the integral. The parallel axis theorem states that the moment of inertia about an axis parallel to the original axis is equal to the moment of inertia about the original axis plus the product of the mass and the square of the distance between the two axes. In this case, we can use the zCOM we found in part (ii) as the new axis. So, Iz = IzCOM + Md^2, where d is the distance between the z-axis and zCOM. Using this, we can simplify the integral to Iz = ∫∫∫l'^2dM, where l' is the perpendicular distance from the zCOM axis. This can be further simplified to Iz = ∫∫∫(rcosθ - zCOM)^2σdV. From here, you can use the same approach as in part (ii) to evaluate the integral and you should get Iz = (4/5)πσ0L^5 as the final answer.

I hope this helps! Keep up the good work.