In cases of rigid body rotation, rotating about axis other than centre of mass we always consider force by gravity completely of body on side of centre of mass why is that so?
Example consider rod of mass M and length L pivoted to wall at distance L/4 from one end then when in horizontal position we will always take torque=MgL/4. Why are we considering total mass on one side when some is on other why dont we use Mg/4 and 3Mg/4 on each sides? I mean what actually occur in system?, if centre of gravity moves out of centre of mass this approximation will be useless.
Newtonian Mechanics – Why Total Gravitational Force Acts on Center of Mass in Rotational Motion
forcesnewtonian-mechanicsreference framesrotational-dynamics
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Your initial calculations are correct. The pin force is indeed $\tfrac{m g}{4} $ a fact that kind of surprised me the first time I encountered this problem.
Your idealization on the second part is where things were missed. I am using the sketch below, and I am counting positive directions as downwards (same as gravity) and positive angles as clock-wise. Notice each half-bar has mass $m/2$ and mass moment of inertia about its center of mass $ \tfrac{1}{12} \left( \tfrac{m}{2} \right) \left( \tfrac{\ell}{2} \right)^2 = \tfrac{m \ell^2}{96}$
Let's look at the equations of motion for the two half-bars as they are derived from the free body diagrams.
$$ \begin{aligned} \tfrac{m}{2} a_G & = \tfrac{m}{2} g - F_C - F_A \\ \tfrac{m \ell^2}{96} \alpha & = -\tau_C - \tfrac{\ell}{4} F_C + \tfrac{\ell}{4} F_A \\ \tfrac{m}{2} a_H & = \tfrac{m}{2} g + F_C \\ \tfrac{m \ell^2}{96} \alpha & = \tau_C - \tfrac{\ell}{4} F_C \\ \end{aligned} $$
And consider the kinematics, where it all acts like a rotating rigid bar, with point accelerations $a_G = \tfrac{\ell}{4} \alpha$ and $a_H = \tfrac{3 \ell}{4} \alpha$
The solution to the above 4×4 system of equations is
$$ \begin{aligned} F_A & = \tfrac{m g}{4} & F_C & = \tfrac{m g}{16} \\ \alpha & = \tfrac{3 g}{2 \ell} & \tau_C &= \tfrac{m g \ell}{32} \end{aligned} $$
I think because you did not account for the torque transfer $\tau_C$ between the half-bars, you got $F_C = \tfrac{m g}{8}$ which is incorrect.
Note, I used PowerPoint and IguanaTex plugin for the sketches.
Best Answer
If a point $P_0$ is the center of mass, we have by definition: $m_1\mathbf {r_1} + m_2\mathbf {r_2} + ... + m_n\mathbf {r_n} = 0$
Taking the derivative with respect to time: $m_1\mathbf {v_1} + m_2\mathbf {v_2} + ... + m_n\mathbf {v_n} = 0$. So the momentum of the body for a non rotating frame attached to the COM is always zero. If there are no external forces applied, its total momentum must be constant for any inertial frame:
$\mathbf p_{tot} = m_1(\mathbf {v_1 + u}) + m_2(\mathbf {v_2 + u}) + ... + m_n(\mathbf {v_n + u}) = \sum m_i\mathbf v_i + \sum m_i\mathbf u = 0 + M\mathbf u$
where $\mathbf u$ is the velocity of the COM with respect to the inertial frame, and $M$ the total mass of the body. If there is an uniform gravitational field, with acceleration $\mathbf g$
$$\mathbf F = \frac{d\mathbf p_{tot}}{dt} = M\frac{d\mathbf u}{dt} = M\mathbf g$$