Torque on Non-Uniform Sphere

Classical Mechanics Level pending

A solid non-uniform spherical ball has a radius of 1 1 and is centered on the origin. Its volume-mass-density is:

ρ ( x , y , z ) = x 2 + 2 y 2 + 3 z 2 \rho(x,y,z) = x^2 + 2 y^2 + 3 z^2

There is a particle of mass 100 100 at position ( x , y , z ) = ( 1 , 2 , 1 ) (x,y,z) = (1,2,1) . There is a torque on the ball with respect to the origin, as a result of the gravitational force from the particle.

Determine the absolute value of this torque.

Note: Universal gravitational constant G = 1 G = 1


The answer is 2.79.

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1 solution

Karan Chatrath
Jun 10, 2020

Laying out steps of evaluation. Location of point mass:

r c = ( 1 , 2 , 1 ) \vec{r}_c = (1,2,1)

Location of any point on the sphere in spherical coordinates:

r p = ( r sin θ cos ϕ , r sin θ sin ϕ , r cos θ ) \vec{r}_p = (r \sin{\theta} \cos{\phi},r \sin{\theta} \sin{\phi},r \cos{\theta} )

Force acting on a volume mass element near the point on the sphere:

d F = G ρ m d V ( r c r p ) r c r p 3 d\vec{F} = \frac{G \rho m \ dV \left(\vec{r}_c-\vec{r}_p\right)}{\lvert \vec{r}_c-\vec{r}_p \rvert^3}

The density is written in term of spherical coordinates and:

d V = r 2 sin θ d r d θ d ϕ dV = r^2 \sin{\theta} \ dr \ d\theta \ d\phi

The torque due to this elementary force about the origin is:

d τ = r p × d F d\vec{\tau} = \vec{r}_p \times d\vec{F}

Substituting all expressions and simplifying:

d τ = d τ x i ^ + d τ y j ^ + d τ z k ^ d\vec{\tau} = d\tau_x \ \hat{i} + d\tau_y \ \hat{j} + d\tau_z \ \hat{k} d τ = ( f x i ^ + f y j ^ + f z k ^ ) d r d θ d ϕ d \vec{\tau} = \left(f_x \ \hat{i} + fy \ \hat{j} + f_z \ \hat{k}\right) dr \ d\theta \ d\phi

τ = ( 0 1 0 π 0 2 π f x d r d θ d ϕ ) i ^ + ( 0 1 0 π 0 2 π f y d r d θ d ϕ ) j ^ + ( 0 1 0 π 0 2 π f z d r d θ d ϕ ) k ^ \tau = \left(\int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi} f_x \ dr \ d\theta \ d\phi\right) \ \hat{i} + \left(\int_{0}^{1}\int_{0}^{\pi} \int_{0}^{2\pi} f_y \ dr \ d\theta \ d\phi\right) \ \hat{j} + \left(\int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi}f_z \ dr \ d\theta \ d\phi\right) \ \hat{k} τ = τ x i ^ + τ y j ^ + τ z k ^ \tau = \tau_x \ \hat{i} + \tau_y \ \hat{j} + \tau_z \ \hat{k}

The required answer is:

τ = τ x 2 + τ y 2 + τ z 2 \lvert \tau \rvert = \sqrt{\tau_x^2 + \tau_y^2 + \tau_z^2}

The numerical integration is very time-consuming as it requires high numerical resolution for decent convergence. I consider myself lucky to have got the answer.

Indeed, in my highest resolution simulation, I divided it into a billion pieces N = 1000 , N 3 = 1 0 9 N = 1000, N^3 = 10^9 .

Steven Chase - 1 year ago

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