Question 17: Force on T-Junction Ball
Solution
1. Moment of Inertia ($I$):
The system pivots about the junction. There are three identical masses $m$ at distance $l$ from the pivot.
$$ I = 3ml^2 $$
2. Torque ($\tau$):
Only the middle rod’s mass contributes to the torque about the pivot because the center of mass of the top two balls lies exactly on the pivot (due to symmetry).
$$ \tau = mg l \sin \theta $$
3. Angular Acceleration ($\alpha$):
Using $\tau = I\alpha$:
$$ mg l \sin \theta = (3ml^2) \alpha $$
$$ \alpha = \frac{g \sin \theta}{3l} $$
4. Force Calculation on Middle Ball:
Let forces exerted by the rod be $F_r$ (radial) and $F_t$ (tangential).
- Radial Direction: No radial acceleration initially ($\omega=0$). $$ F_r + mg \cos \theta = 0 \implies F_r = -mg \cos \theta $$ (Force is Tension)
- Tangential Direction: $$ F_t + mg \sin \theta = m a_t = m (l \alpha) $$ $$ F_t + mg \sin \theta = m l \left( \frac{g \sin \theta}{3l} \right) = \frac{mg \sin \theta}{3} $$ $$ F_t = -\frac{2}{3} mg \sin \theta $$
5. Resultant Force:
$$ |F| = \sqrt{F_r^2 + F_t^2} = mg \sqrt{\cos^2 \theta + \frac{4}{9}\sin^2 \theta} $$
Substituting $m=\sqrt{5}$, $g=10$, $\theta=37^\circ$:
$$ F = 20 \text{ N} $$
Angle with rod: $\beta = \tan^{-1} \left( \frac{F_t}{F_r} \right) = \tan^{-1}\left( \frac{1}{2} \right)$.