RBD CYU 9

Let $O$ be the center of the cylinder and $r$ be its radius. The rod touches the cylinder at $P$ and the cord leaves the rod at tip $A$ and touches the cylinder at $S$. Since $AP$ and $AS$ are tangent segments from point $A$ to the circle: $$ AP = AS = \eta r $$ Let $\angle POA = \alpha$. In the right-angled triangle $\Delta OPA$: $$ \tan \alpha = \frac{AP}{OP} = \frac{\eta r}{r} = \eta $$ Given $\eta = 1/\sqrt{2}$, we have $\tan \alpha = 1/\sqrt{2}$. Also, $\Delta OPA \cong \Delta OSA$, so $\angle SOA = \alpha$. The total angle between the two contact radii is $\angle POS = 2\alpha$. The angle between the two tangents $AP$ (line of the rod) and $AS$ (line of tension) is related to the central angle. Specifically, the angle $\angle PAS = 180^\circ – 2\alpha$.

The cylinder is frictionless, so the reaction force $N$ at $P$ is purely normal (perpendicular) to the rod. For the rod to remain in equilibrium (no sliding), the net force parallel to the rod must be zero. Forces parallel to the rod: 1. Component of Weight ($Mg$): Acts downwards. Component along rod slope = $Mg \sin \theta$ (down the slope). 2. Component of Tension ($T$): Tension acts along $AS$. Let’s find the component of $T$ along the rod (up the slope). The angle between the vector $\vec{AP}$ (pointing down slope) and tension vector $\vec{AS}$ is $180^\circ – 2\alpha$. Therefore, the projection of $T$ onto the line of the rod is: $$ T_{\parallel} = T \cos(180^\circ – 2\alpha) = -T \cos(2\alpha) $$ This negative sign indicates the projection points opposite to $\vec{AP}$, i.e., up the slope. Force Balance Equation: $$ Mg \sin \theta = T \cos(2\alpha) $$ Since the load is $m$, Tension $T = mg$. $$ Mg \sin \theta = mg \cos(2\alpha) $$ Using $\tan \alpha = 1/\sqrt{2}$: $$ \cos(2\alpha) = \frac{1 – \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{1 – 0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{1}{3} $$ Substitute values ($M=10, \theta=30^\circ$): $$ 10 \cdot \sin(30^\circ) = m \cdot \frac{1}{3} $$ $$ 10 \cdot \frac{1}{2} = \frac{m}{3} \implies 5 = \frac{m}{3} $$ $$ m = 15 \text{ kg} $$

Now apply torque balance about the point of contact $P$. 1. Torque of Weight: The center of mass $G$ is at distance $l/2$ from $A$. The contact $P$ is at distance $\eta r$ from $A$. Distance $PG = l/2 – \eta r$. Lever arm horizontal projection = $(l/2 – \eta r) \cos \theta$. $$ \tau_{Mg} = Mg \left( \frac{l}{2} – \eta r \right) \cos \theta $$ 2. Torque of Tension: Force $T$ acts at $A$. The lever arm is the perpendicular distance from $P$ to line $AS$. In $\Delta PAS$, sides $PA=AS=\eta r$. The angle at $A$ is $180^\circ – 2\alpha$. Perpendicular distance $h = AP \sin(\angle PAS) = \eta r \sin(180^\circ – 2\alpha) = \eta r \sin(2\alpha)$. $$ \tau_{T} = T (\eta r \sin 2\alpha) $$ Equating Torques: $$ Mg \left( \frac{l}{2} – \eta r \right) \cos \theta = T \eta r \sin 2\alpha $$ Substitute $T = \frac{Mg \sin \theta}{\cos 2\alpha}$ (from Force Balance step): $$ Mg \left( \frac{l}{2} – \eta r \right) \cos \theta = \left( \frac{Mg \sin \theta}{\cos 2\alpha} \right) \eta r \sin 2\alpha $$ Cancel $Mg$ and simplify using $\tan 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha}$: $$ \left( \frac{l}{2} – \eta r \right) \cos \theta = \eta r \sin \theta \tan 2\alpha $$ $$ \frac{l}{2} – \eta r = \eta r \tan \theta \tan 2\alpha $$ $$ \frac{l}{2} = \eta r ( 1 + \tan \theta \tan 2\alpha ) $$ $$ \frac{l}{r} = 2\eta ( 1 + \tan \theta \tan 2\alpha ) $$