Maximum Excursion (Part 2)

Classical Mechanics Level 5

In the $xy$ -plane, there are two massive particles. Particle 1 $($ mass $m_1 = 1\text{ kg})$ is attached to a frictionless, massless wire in the shape of the curve $y = \alpha x^{2}$ , where $(\alpha = 1 \, m^{-1})$ . It is free to slide along the wire, and is initially at rest at $(x,y) = (-1\text{ m},1\text{ m})$ .

Particle 2 is fixed at $(x,y) = (-1\text{ m},0\text{ m})$ .

There is a uniform ambient downward gravitational acceleration of $10\text{ m/s}^{2}$ .

Particle 1 slides down the wire under the influence of the gravitational pull of Particle 2 and of the ambient gravity. If Particle 1 reaches $\big( x = +\frac{1}{2}\text{ m}\big)$ before stopping and coming back, determine the mass of Particle 2 (see notes below).

Details and Assumptions:

The universal gravitational constant is $\SI[per-mode=symbol]{6.674e-11}{\newton\meter\squared\per\kilo\gram\squared}$ .
Report your answer as the mass of Particle 2, in multiples of $10^{11}\text{ kg}$ , to 2 decimal places.

The answer is 3.28.

1 solution

Steven Chase
Nov 21, 2016

The particle stops moving at $(x,y) = \left(\frac{1}{2},\frac{1}{4}\right)$ . There are two potential energy expressions to consider. One is the interaction of the bead on the wire with the ambient field $(m_1 g y = m_1 g \alpha x^{2})$ and the other is the interaction between the masses $(\frac{-Gm_1m_2}{r})$ .
The particle doesn't rise to its original height on the positive- $x$ side because some of the initial potential energy is going into further separating particles 1 and 2.

The potential energy balance equation is:

$\begin{aligned} m_1g(y_i - y_f) &= Gm_1m_2\left(\frac{1}{d_i} - \frac{1}{d_f}\right) \\ m_1 g \alpha\left(1 - \frac{1}{4}\right) &= Gm_1m_2 \left(\frac{1}{1} - \frac{1}{\sqrt{(\frac{3}{2})^{2} + (\frac{1}{4})^{2}}}\right). \end{aligned}$

$m_1$ appears on both sides and cancels, allowing us to solve for $m_2$ . $m_2 \approx 3.28 \times 10^{11} kg$

s/charges/masses/

Shaun Griffith - 4 years, 6 months ago

Edited, thanks

Steven Chase - 4 years, 6 months ago

hey, i did the exact same thing, just exact , but i was getting answer as 2.78 , i typed in wolfram ... 3 10/(4 6.67 (1-4/(3 sqrt5)) see it once !

A Former Brilliant Member - 4 years, 5 months ago

Maximum Excursion (Part 2)

The answer is 3.28.

1 solution

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