RBD BYU 20

Note: The Normal Reaction on the plates has a vertical component which balances the weight of the plates.

1. Torque Equation

Consider the torque about the hinge. The torque due to the weight of the rod (acting at $l/2$) is balanced by the torque due to the normal reaction force $\mathcal{N}$ from the cylinder (acting at distance $x$): $$ mg \frac{l}{2} \sin\theta = \mathcal{N} x \quad \text{— (i)} $$

2. Geometric Derivation of $x$

From the diagram, consider the right-angled triangle formed by the hinge, the center of the cylinder, and the point of contact. The radius $R$ is perpendicular to the tangent (plate). $$ \tan \theta = \frac{R}{x} \implies x = R \cot \theta = R \frac{\cos \theta}{\sin \theta} \quad \text{— (ii)} $$

3. Force Relation

Using the condition given in the note (vertical component of normal reaction balances weight): $$ \mathcal{N} \sin \theta = mg \implies \mathcal{N} = \frac{mg}{\sin \theta} \quad \text{— (iii)} $$

4. Combining the Equations

Substitute values of $x$ from (ii) and $\mathcal{N}$ from (iii) into the torque equation (i): $$ mg \frac{l}{2} \sin \theta = \left( \frac{mg}{\sin \theta} \right) \left( R \frac{\cos \theta}{\sin \theta} \right) $$ Canceling $mg$ from both sides: $$ \frac{l}{2} \sin \theta = \frac{R \cos \theta}{\sin^2 \theta} $$ Rearranging the terms: $$ l \sin^3 \theta = 2 R \cos \theta $$

5. Solving for $\theta$

Rearranging the derived relationship: $$ 2 \frac{\sin^3 \theta}{\cos^3 \theta} = \frac{\cos \theta}{\cos^3 \theta} $$ $$ 2 \tan^3 \theta = \sec^2 \theta $$ Using $\sec^2 \theta = 1 + \tan^2 \theta$ and letting $u = \tan\theta$: $$ 2u^3 = 1 + u^2 \implies 2u^3 – u^2 – 1 = 0 $$ Factorizing the polynomial (trial solution $u=1$): $$ (u-1)(2u^2 + u + 1) = 0 $$ Since $2u^2 + u + 1 = 0$ has no real roots, the only solution is $u = 1$. $$ \tan \theta = 1 \implies \theta = 45^\circ $$