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For rotational equilibrium of a rigid object, about which point does the torque of the body have to be zero?

The torque has to be zero about all points -- if it was nonzero for some point, the object would start to rotate around that point. On the other hand, you often see that the torque is calculated ...
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For rotational equilibrium of a rigid object, about which point does the torque of the body have to be zero?

For rotational equilibrium of an object, the sum of the torques about any point on the object must be zero. Hope this helps.
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Can we say torque at each point is same

This appears to be a homework problem so I will not provide a full answer. I assume the rod in equilibrium (not rotating or translating). If so, the sum of torques is zero. Given the torque from F ...
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How can you take torque about an accelerating point that isn't the center of mass?

At the instant of maximum displacement, the point of attachment is instantaneously at rest. That point in space can be taken as a point of reference for the rate of change of the angular momentum of ...
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How can you take torque about an accelerating point that isn't the center of mass?

Here is the full concept. You can find torque about any point. Problem is you cannot apply $\tau = I \alpha$ about any point. To apply $\tau = I \alpha$, we need to make sure that the point is ...
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Why is torque defined as $\vec{r} \times F$?

Torque is change of angular momentum: $$\vec{\tau} = \frac{d\vec{L}}{dt}$$ Angular momentum is defined as $$\vec{L} = \vec{r} \times\vec{p}$$ Using the chain rule:  \vec{\tau} = \frac{d\vec{L}}{...
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I think I found a simpler way. First you prove $(\sum_i{\tau_i})\Delta\theta$ is the change in total kinetic energy for a small $\Delta\theta$. You know the change in a single particle's kinetic ...
The General Case Let $k$ and $j$ be any two particles, and $\vec F^i_{k\to j}$ be the (internal) force of particle $k$ on $j$. For the total internal torque to vanish, a stronger form of Newton's ...