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6

Like Wikipedia says: "Moment is a combination of a physical quantity and a distance." This 'physical quantity' could be various things. To take the examples you mention: Moment of momentum (commonly known as angular momentum) is expressed as $\vec{L}=\vec{r}\times m\vec{v}$, and is a measure for the rotational momentum of an object around some axis. Moment ...


3

It is due to the viscous nature of any liquid. When you stir, the liquid starts spinning and this causes the liquid (the part which is in contact with the pot) to "drag" the pot(due to friction) along with it in the path of its motion.Hope this answers your question.


2

When you don't have to consider the inertia of the screw (which is always), these problems can easily be analyzed by unwinding the screw to a plane (as you've done in your figure). So think of the screw and the surface of contact as two inclined planes that slide against each other (one on top and one beneath it). Now, pushing on the screw is like pushing ...


2

The difference is only how you define $\theta$ and the zero of potential energy. The $\cos \theta$ expression takes the zero of potential energy to be when $\theta = \frac \pi 2$ whereas you derivation with $\sin \theta$ in it takes the zero of potential to be when $\theta = 0$.


1

Must admit this got me really worried, and I wondered if the problem was maybe in the use of $\bf{\tau} = \bf{r \times F}$ and identifying the force with the (negative) gradient of some potential. Then, I really got worried because it seemed to me that the same problem would occur if you were to make similar calculations with determining the torque on an ...


1

A rigid body moving with no constraints, in particular rotating, will rotate necessarily about a principal axis of inertia. I thought that the reason of this is that otherwise, the angular momentum $\vec L$ would not be parallel to the angular velocity $\vec\omega$, hence it would follow a precession motion and that would imply the presence of a ...


1

You have to approach problems systematically, and not intuitively. Like I stated in a previous (accepted) answer, resolve everything on the center of mass, and only in the end transfer the quantities to a different point (like P) to get the results you want. I start with the kinematics. Use $\ell_1$ and $\ell_2$ for the horizontal distances and $h$ for the ...


1

This is a question in solid mechanics. Under the action of a constant torque along its length, the shaft experiences a deformation, such that one end of the shaft is rotated slighly relative to the other. But it is not enough to cause the shaft to warp, and the shaft returns elastically to its original shape after the torque is removed. Each cross section ...


1

Hint: first find the angular velocity of the disc by conservation of energy, $\frac{1}{2}I\omega^2 = mgh$ , $\omega$ = angular velocity of the pulley, I = moment of inertia of the pulley. It is mentioned that the pendulum cover a horizontal distance L in t seconds. And we also know that $$v_{linear-velocity} = R\omega$$ So time t can be found ...


1

A probable explanation for this effect is simply that the bottom of the pot might be a bit bulged out, as to form only one point of contact around which the pot then can rotate relatively freely (with little friction). As you stir the water inside the pot, the moving water molecules exert a frictional force on the walls of the pot, dragging it in the same ...


1

Gareth, this is a good first question. You've asked it well, and I encourage you to ask more questions in the future. Basically, you can just reverse any usual explanation of a screw. Normally, the wedge transfers torque into thrust; well in this case the wedge transfers thrust into torque — it's exactly the same, but reversed. This is a special example ...


1

Place a coordinate system at O oriented such that $\boldsymbol{r}_i = (x_i,0,z_i)$. The body is rotating with $\boldsymbol{\omega} = (0,0,\Omega)$ and accelerating with $\boldsymbol{\alpha} = (0,0,\dot\Omega)$ about the z axis, and the mass mi lies on the xz plane. Then linear momentum of mi is $$ \boldsymbol{p}_i = -m_i \boldsymbol{r}_i \times ...



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