Question 25: Rigid Rod Acceleration (Alternative Method)
Problem: A light rigid rod connects masses $m$ and $3m$. A force $F$ is applied at the midpoint. Find the acceleration of the midpoint.
1. Force Analysis on Light Rod
Since the rod is light (massless), the net force on the rod and the net torque about its center of mass must be zero.
- Force Balance: $F – T_1 – T_2 = 0$
- Torque Balance (about midpoint): $T_1(\frac{l}{2}) = T_2(\frac{l}{2}) \implies T_1 = T_2$
Substituting back:
$$ 2T_1 = F \implies T_1 = T_2 = \frac{F}{2} $$2. Acceleration of Blocks
Using Newton’s Second Law for each mass:
$$ a_1 = \frac{T_1}{m} = \frac{F/2}{m} = \frac{F}{2m} $$ $$ a_2 = \frac{T_2}{3m} = \frac{F/2}{3m} = \frac{F}{6m} $$3. Acceleration of Rod’s Center
For a rigid rod undergoing planar motion, the acceleration of the midpoint is the average of the accelerations of the ends (linear relation).
$$ a_{mid} = \frac{a_1 + a_2}{2} $$ $$ a_{mid} = \frac{\frac{F}{2m} + \frac{F}{6m}}{2} $$ $$ a_{mid} = \frac{\frac{3F}{6m} + \frac{F}{6m}}{2} = \frac{\frac{4F}{6m}}{2} $$ $$ a_{mid} = \frac{4F}{12m} = \frac{F}{3m} $$
$$ a_{mid} = \frac{F}{3m} $$