Wavy Ring Gravity

Classical Mechanics Level 3

A wavy ring is parametrized below:

x = cos ⁡ θ y = sin ⁡ θ z = sin ⁡ θ 0 ≤ θ ≤ 2 π x = \cos \theta \\ y = \sin \theta \\ z = \sin \theta \\ 0 \leq \theta \leq 2 \pi

The ring has σ \sigma units of mass per unit curve length. There is a point particle of mass m m positioned at point T ⃗ \vec{T} .

What is the magnitude of the gravitational force on the point particle?

Details and Assumptions:
1) σ = 1 \sigma = 1
2) m = 1 m = 1
3) Universal gravity constant G = 1 G = 1 (for simplicity). There is no ambient gravity
4) T ⃗ = ( T x , T y , T z ) = ( 0 , 0 , 1 ) \vec{T} = (T_x,T_y,T_z) = (0,0,1)


The answer is 2.536.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Jun 5, 2019

Let the position vector of a point on the ring be:

r ⃗ = x i ^ + y j ^ + z k ^ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

or:

r ⃗ = cos ⁡ ( θ ) i ^ + sin ⁡ ( θ ) j ^ + sin ⁡ ( θ ) k ^ \vec{r} = \cos(\theta) \hat{i} + \sin(\theta) \hat{j} + \sin(\theta) \hat{k}

The position vector of the point of interest (at which force needs to be computed) is:

r ⃗ p = 0 i ^ + 0 j ^ + 1 k ^ \vec{r}_p = 0 \hat{i} + 0 \hat{j} + 1 \hat{k}

It is given that the mass per unit length of this ring is unity. Let us consider a small strip of wire which is a part of the ring. The length of this strip is the arc element which is computed as such:

d s = ( d x d θ ) 2 + ( d y d θ ) 2 + ( d z d θ ) 2 d θ ds = \sqrt{\bigg(\frac{dx}{d\theta}\bigg)^2 + \bigg(\frac{dy}{d\theta}\bigg)^2 + \bigg(\frac{dz}{d\theta}\bigg)^2} d\theta

Therefore, the mass of this element is:

d M = σ d s dM = \sigma ds

The gravitational force can be computed by using the inverse square law:

d F ⃗ = d F x i ^ + d F y j ^ + d F z k ^ d\vec{F} = dF_x \hat{i} + dF_y \hat{j} + dF_z \hat{k}

or:

d F ⃗ = G ( d M ) m ∣ r ⃗ − r ⃗ p ∣ 2 ( r ⃗ − r ⃗ p ∣ r ⃗ − r ⃗ p ∣ ) d\vec{F} = \frac{G(dM)m}{\mid\vec{r} - \vec{r}_p\mid^2}\bigg(\frac{\vec{r} - \vec{r}_p}{\mid\vec{r} - \vec{r}_p\mid}\bigg)

The expressions for force elements in terms of θ \theta come out to be:

d F x = cos ⁡ ( θ ) cos ⁡ ( θ ) 2 + 1 ( sin ⁡ ( θ ) 2 − 2 sin ⁡ ( θ ) + 2 ) 3 / 2 d θ dF_x = \frac{\cos\left(\mathrm{\theta}\right)\,\sqrt{{\cos\left(\mathrm{\theta}\right)}^2+1}}{{\left({\sin\left(\mathrm{\theta}\right)}^2-2\,\sin\left(\mathrm{\theta}\right)+2\right)}^{3/2}} d\theta

d F y = sin ⁡ ( θ ) 2 − sin ⁡ ( θ ) 2 ( sin ⁡ ( θ ) 2 − 2 sin ⁡ ( θ ) + 2 ) 3 / 2 d θ dF_y = \frac{\sin\left(\mathrm{\theta}\right)\,\sqrt{2-{\sin\left(\mathrm{\theta}\right)}^2}}{{\left({\sin\left(\mathrm{\theta}\right)}^2-2\,\sin\left(\mathrm{\theta}\right)+2\right)}^{3/2}}d\theta

d F z = ( sin ⁡ ( θ ) − 1 ) 2 − sin ⁡ ( θ ) 2 ( sin ⁡ ( θ ) 2 − 2 sin ⁡ ( θ ) + 2 ) 3 / 2 d θ dF_z = \frac{\left(\sin\left(\mathrm{\theta}\right)-1\right)\,\sqrt{2-{\sin\left(\mathrm{\theta}\right)}^2}}{{\left({\sin\left(\mathrm{\theta}\right)}^2-2\,\sin\left(\mathrm{\theta}\right)+2\right)}^{3/2}}d\theta

Each of these expressions need to be integrated from 0 0 to 2 π 2\pi .

The integrals are computed by breaking them into parts and numerically integrating: F x = ∫ 0 π / 2 d F x + ∫ π / 2 π d F x + ∫ π 3 π / 2 d F x + ∫ 3 π / 2 2 π d F x Fx = \int_{0}^{\pi/2}dF_x + \int_{\pi/2}^{\pi}dF_x + \int_{\pi}^{3\pi/2}dF_x + \int_{3\pi/2}^{2\pi}dF_x

Same can be done for the Y and Z components.

The required answer is:

F = F x 2 + F y 2 + F z 2 F = \sqrt{F_x^2 + F_y^2 + F_z^2}

Which comes out to be: 2.536

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