Piece-wise Uniform Rod Equilibrium

Classical Mechanics Level 3

Consider a uniform rod of length $b$ . One half of this rod has a linear mass density of $\lambda_1 = 9$ while the other half has a linear mass density of $\lambda_2 = 1$ . The rod is initially placed horizontally in a smooth, fixed hemispherical bowl and the denser side is on the left. It is then released from rest at this initial position. The rod undergoes some motion but eventually comes to rest due to the presence of other dissipative forces. The angle $\theta$ that the rod makes with the horizontal when it is in equilibrium is given by:

$\tan{\theta} = \frac{m}{n} \left(\frac{1}{\sqrt{\left(\frac{pR}{b}\right)^q -1}}\right)$

Here, $m,\ n,\ p,\ q$ are positive integers and $m$ and $n$ are co-prime. FInd $\boxed{mnpq}$ .

Note:

This problem is not original.
The diagram on the left shows the initial configuration.
Gravity acts vertically downwards.

The answer is 40.

2 solutions

Mark Hennings
Oct 11, 2019

It is an easy calculation that the centre of mass $G$ of the rod is such that $AG = \tfrac{3}{10}b$ .

In equilibrium, the three forces acting on the rod (gravity and the two normal reactions) must have concurrent lines of action. Thus the centre of mass $G$ must be vertically beneath the centre $O$ of the circle. Thus $\begin{aligned} R \cos\left(\theta + \cos^{-1}\tfrac{b}{2R}\right) & = \; \tfrac{3}{10}b\cos\theta \\ \tan\theta & = \; \frac25\left(\left(\frac{2R}{b}\right)^2 - 1\right)^{-\frac12} \end{aligned}$ so that $m=2$ , $n=5$ , $p=q=2$ , making the answer $\boxed{40}$ .

A Former Brilliant Member
Oct 11, 2019

Solving the force balance equations along the horizontal and vertical directions, and the moment balance equation about the lower end of the rod, we get $\tan \theta=\dfrac{b\times (\lambda_1-\lambda_2)}{2\times (\lambda_1+\lambda_2)\times \sqrt {4R^2-b^2}}$ . So $m=2,n=5,p=2,q=2$ and $\boxed {mnpq=40}$

Piece-wise Uniform Rod Equilibrium

The answer is 40.

2 solutions

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