Question

A thin uniform rod of mass m and length $\ell$ is free to rotate about a horizontal axis passing through end A. The rod is released from rest when it is horizontal at time t = 0. $a_B$ and $a_C$ are accelerations of end point B and mid-point C respectively. $F_H$ is the force by the hinge on the rod (g = 10 m/s²) and $\theta$ is the angular displacement from initial position.

• A. At t=0, $a_B = \frac{3g}{2}$, $a_C = \frac{3g}{4}$, $F_H = \frac{mg}{4}$
• B. At t=0, $a_B = \frac{3g}{2}$, $a_C = \frac{3g}{2}$, $F_H = \frac{mg}{4}$
• C. At t=0, $a_B = \frac{3g}{4}$, $a_C = \frac{3g}{4}$, $F_H = \frac{mg}{4}$
• D. At t=0, $a_B = \frac{3g}{2}$, $a_C = \frac{3g}{4}$, $F_H = \frac{3mg}{4}

Answer 1

Answer: (A)

At t=0, $a_B = \frac{3g}{2}$, $a_C = \frac{3g}{4}$, $F_H = \frac{mg}{4}

Answer 2

1. Calculate the initial angular acceleration using the torque due to gravity about the pivot point A.
   $\tau = I\alpha \implies mg\frac{l}{2} = \frac{ml^2}{3}\alpha \implies \alpha = \frac{3g}{2l}$

2. Calculate the initial linear acceleration of points B and C using the relation $a = r\alpha$ for tangential acceleration, noting that the radial acceleration is zero initially since $\omega=0$.
   $a_B = l\alpha = \frac{3g}{2}$
   $a_C = \frac{l}{2}\alpha = \frac{3g}{4}$

3. Calculate the initial hinge force by applying Newton's second law to the center of mass and relating its acceleration to the angular acceleration.
   $F_{net} = ma_{cm} \implies F_H - mg = -m(\frac{3g}{4}) \implies F_H = \frac{mg}{4}$