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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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Compute angular acceleration from torque in 3D

The equations of motion for rotation about the center of mass are $$ \boldsymbol{\tau} = \mathbf{I}\,\boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathbf{I} \boldsymbol{\omega} \tag{1} $$ where $\...
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A dilemma regarding torque when a body moves in circular motion

The example is similar to the olympic hammer throw. The athlete increases the tangential speed of the hammer by increasing the angular speed of his hands around his body. That is: the center of ...
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A dilemma regarding torque when a body moves in circular motion

Angular acceleration is the rate of change of the rate of angular velocity (relative to the point that the angle is measured from). If the string revolves with constant angular velocity it has no ...
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A dilemma regarding torque when a body moves in circular motion

"Now the mass 'm' accelerates in the circular path and thus there is an angular acceleration and thus tangential acceleration" The problem starts from above . The mass m accelerates as it's ...
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A dilemma regarding torque when a body moves in circular motion

Now the mass 'm' accelerates in the circular path and thus there is an angular acceleration and thus tangential acceleration. Your question is very unclear. Masses don’t just accelerate. There is a ...
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When two systems of forces acting on a rigid body are equivalent?

The equations of motion can be thought of given a specified motion, provided by the translation acceleration vector of the center of mass $\boldsymbol{a}_C$ and the rotation acceleration vector of the ...
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3 votes

When two systems of forces acting on a rigid body are equivalent?

A theorem for a rigid body is: "Every system of forces is equivalent to a single force through an arbitrary point, plus a couple (either or both of which may be zero.)". A couple is defined ...
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3 votes

What rotates a disk hung at center of mass when it is tilted?

Most methods of attachment bend the string when you tilt the plate at the point where the string connects to the plate. Strings under tension want to be straight. Along with non-zero plate thickness ...
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3 votes

Torque due to magnetic field

$\overrightarrow{\tau} = \overrightarrow{M} X \overrightarrow{B}$ is valid only in uniform magnetic field. Any point can be taken as origin and torque will be same. To prove it, first of all prove it ...
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Do we take torque/angular momentum about a point or an axis?

In general, torque and angular momentum are taken about a point. For the special case of rotation of a rigid body about a fixed axis (or more correctly rotation in a plane if the axis moves, such a ...
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This question is about motor power requirements

The 2nd question is not answered above. Power is 'rate of doing work' and in this case is 'rate of increase of KE' (assuming no losses!) Now - KE = 0.5 mv^2 and v= at (for uniform acceleration and ...
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This question is about motor power requirements

Question one: yes indeed, both conditions you describe represent the same power. Now, why choose one over another? Well, some applications work best with high speed and low torque (like a wood-cutting ...
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Why is torque defined as $\vec{r} \times F$?

I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically ...
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10 votes

Why is torque defined as $\vec{r} \times F$?

It's not like someone said "Ah ha, torque! What should the definition of torque be?" That doesn't make sense; you don't think up a term and then try to assign a definition. Instead, it was ...
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2 votes

Why is torque defined as $\vec{r} \times F$?

Consider a point particle of mass $m$ with velocity $\vec v$. The particle is located at some position $\vec r$ with respect to the origin $O$. I will start with the angular momentum calculated about $...
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4 votes

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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Why is the sum of torques for each particle equal to the external torque?

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 ...
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Why do internal torques cancel?

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 ...
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