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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.

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  • $\begingroup$ This isn't an answer to your question, but it may help you think about rotational motion and how considering forces on little pieces of an object can lead to a simple view of how they act on the whole object. Toppling of a cylinder on a block $\endgroup$
    – mmesser314
    Dec 6, 2021 at 15:37

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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 $\mathbf g$

$$\mathbf F = \frac{d\mathbf p_{tot}}{dt} = M\frac{d\mathbf u}{dt} = M\mathbf g$$

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Because a rigid body acts like its entire mass is contained within a point called center of mass. This greatly simplifies solving problems such as those in your examples. But in principle you can adopt "your approach" given in your example, where you consider that both sides have mass, although this unnecessarily complicates the solution. For example, you could say that forces acting on two sides of your rod are $mg/4$ and $3mg/4$. Now, the distances of those forces to the rotational axis of the rod are $l/8$ and $3l/8$ respectively. And you can see that they give opposite torques, meaning that the total torques is $3mg/4\cdot 3l/8 - mg/4\cdot l/8=mgl/4$, which is the same as you would get if you consider that the entire mass of the rod is contained in its center of mass.

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  • $\begingroup$ Yes but what actually occurs in system $\endgroup$
    – UV0
    Dec 6, 2021 at 12:44
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    $\begingroup$ What happens is that force of gravity acts on every small segment of your rod so total torque would be the sum of torques produced on every segment of your rod. That's hard to calculate though (well it isn't that hard if you know integrals), but luckily you can prove mathematically that the system (i.e. rod) behaves as if the force of gravity only acts on its center of mass which makes the problem much easier to solve. For example, try to understand Claudio's proof in the answers where he demonstrates that you can take that gravity acts only on body's center of mass (if g is uniform in space). $\endgroup$ Dec 7, 2021 at 10:12
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The center of gravity of a system lies on a vertical line through the system about which gravity produces no net torque. To locate it as a point within a solid object, you have to rotate the object and find a new vertical line. Then the center is where the two lines meet. The center of gravity coincides with the center of mass only in a uniform gravitational field.

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The center of mass is a special point that decouples the equations of motion between translations and rotations.

Considering only gravity and no other effect, applying the weight through the center of mass is sufficient to completely describe the motion of the object. This is a result of the fact that translational momentum can be completely defined by the product of the total mass with the motion of the center of mass. And by Newton's 2nd law net force equals the change in translational momentum.

So applying the weight through the center of mass (or in fact resolving the equations of motion at the center of mass) is a result of the definition of momentum, Newton's 2nd law, and what makes things easier to solve.

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