RBD CYU 16

Question 16: Tension in Triangular Structure

Solution Analysis

1. Derivation of Angular Velocity ($\omega$):
The structure is rigid, so the distance between any two balls is constant. This implies the relative velocity between any two balls along the line joining them must be zero.
Let velocity of A be $v$ along AB. Let velocity of B be $v_B$ along BC. Consider the rod AB.

• Component of $v_A$ along AB: $v$.
• Component of $v_B$ along AB: The angle between vector BC (direction of $v_B$) and vector BA (direction of rod) is $60^\circ$. Wait, the exterior angle is $120^\circ$. $v_B$ is along BC. The rod is AB. The component of $v_B$ along the extension of AB (projection onto AB line) is $v_B \cos(120^\circ) = -v_B/2$.

Equating the components of velocity along the rod AB: $$ v_{along} = v_A = v_{B, along} $$ $$ v = -v_B/2 \implies |v_B| = 2v $$ Now, to find $\omega$, we use the relative velocity perpendicular to the rod AB. $$ \omega = \frac{|v_{BA, \perp}|}{l} $$ $$ v_{BA, \perp} = v_{B, \perp} – v_{A, \perp} $$ $v_A$ is entirely along AB, so $v_{A, \perp} = 0$. $v_B$ component perpendicular to AB: $v_B \sin(60^\circ) = (2v) \frac{\sqrt{3}}{2} = v\sqrt{3}$. $$ \omega = \frac{v\sqrt{3}}{l} $$

2. Force Analysis:
Consider the frame of the centroid. The tension $T$ in the rods provides the centripetal force. Radius of rotation $R = \frac{l}{\sqrt{3}}$. Centripetal force required on mass $m$: $$ F_c = m \omega^2 R = m \left( \frac{\sqrt{3}v}{l} \right)^2 \left( \frac{l}{\sqrt{3}} \right) = \frac{\sqrt{3} mv^2}{l} $$ The resultant of the tensions from the two rods connected to a mass (angle $60^\circ$) is: $$ F_{res} = 2 T \cos(30^\circ) = T\sqrt{3} $$

3. Final Calculation:
$$ T\sqrt{3} = \frac{\sqrt{3} mv^2}{l} $$ $$ T = \frac{mv^2}{l} $$

Answer

$$ T = \frac{mv^2}{l} $$