# Tag Info

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You have made a model of a viscous fluid coupling which was used in a number of four wheel drive vehicles to transfer torque. The system relies in the fact that adjacent planes of moving liquid experience a viscous force between them.

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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.

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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 ...

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The space ship should undergo both linear motion and angular (rotational motion). Linear motion will take place in the direction in which the force is applied, and rotation about its centre of gravity. So the centre of gravity will act as an imaginary pivot.

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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 ...

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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 ...

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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 ...

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$\sum t=dL/dt$ on an inertial system, where L is the angular momentum then if there is a torque there will be a change on the L. In this case L changes direction and because of that trajectory is in approximation a circle. We can say there is a total torque, because of it there is a change on L. If there is a resultant torque, there will be a change on L ...

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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 ...

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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 ...

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If you study the problem in the frame of reference of the CM, the contact force of the plane or the rope does work, because the point of contact is moving in this frame. However, note this frame is not galilean, so you may not blindly apply the work-energy theorem (in this exact case it is valid, but take the good habit not to apply it in non-galilean ...

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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 ...

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Any force whose line of action is not through the centre of mass of a body can be transformed into the same magnitude and direction force acting through the centre of mass and a couple. A couple (two parallel, non-colinear, equal in magnitude but opposite in direction force) has the special properties that its torque is independent of the point about which ...

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Because the system is in equilibrium only in the CM frame. If you calculate the moment about any other point, you need to consider pseudo force as well, viz., 133.4N pointing to the left passing through the CM. Then you will get the correct answer again. If you do not want to work in the CM frame, then you should notice that the angular momentum is $${\bf ... 0 The important factor here is the precise definition of the constraint at O. From your description of what the textbook describes occurring, I take it that the rod is supported at O by a constraint which allows rotation, but only about some axis which is part of the definition of the constraint (a hinge, not a ball joint). Such a constraint can exert ... 0 The axis of rotation changing even when there is no applied torque is an effect which is possible in 3-dimensions but not in 2-dimensions. The angular momentum about an axis \vec L is given by \vec L = I \vec \omega where I is the moment of inertia about the axis and \vec \omega is the angular velocity about the axis. If you are observing an ... 1 A rigid body moving with no constraints, in particular rotating, will rotate necessarily about a principal axis of inertia. I thought that the reason of this is that otherwise, the angular momentum \vec L would not be parallel to the angular velocity \vec\omega, hence it would follow a precession motion and that would imply the presence of a ... 0 The book is wrong and you are correct.$$\begin{pmatrix} -\frac{L}{2} \cos 60° \\ \frac{L}{2} \sin 60° \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ -W \\ 0 \end{pmatrix} + \begin{pmatrix} -L \cos 60° \\ L \sin 60° \\ 0 \end{pmatrix} \times \begin{pmatrix} F_3 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\rightarrow ...

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If you do a free body diagram you will see that a moment needs to be added in order to balance the forces and moments. The forces will be equal and opposite and the moment will depend on the distance between the forces as well as their magnitude. $$M = F d$$

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Here is the beginning of a free body diagram. Each contact force $A$ and $B_y$ acts on the contact normal direction. For $B_y$ this is vertical and for $A$ this is perpendicular to the lever. The weight $W=m g$ acts on center of mass. Now you have to add friction to the contacts as shown with force $B_x$. Here friction is shown with a positive value ...

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Maybe, I found the solution. let's denote with $\theta_k$ the rotation around $k$-axis. Suppose that both $\vec{m}$ and $\vec{B}$ belongs to $xy$ plane. That is: $$\vec{m} = \begin{pmatrix} m \cos(\theta_z)\\ m \sin(\theta_z)\\ 0 \end{pmatrix} ~\text{and} ~\vec{B} = \begin{pmatrix} B \\ 0\\ 0 \end{pmatrix}.$$ The torque using the formula $\vec{\tau} = ... 1 Must admit this got me really worried, and I wondered if the problem was maybe in the use of$\bf{\tau} = \bf{r \times F}$and identifying the force with the (negative) gradient of some potential. Then, I really got worried because it seemed to me that the same problem would occur if you were to make similar calculations with determining the torque on an ... 0 I don't understand why all answerers immediately jump to the topic of momentum, carrying by the field. Nowadays nobody doubt field can carry momentum. The questions like this and especially mine one: How to observe Newton's third law violation? have different aspect on how this is possible, that third law is violated. Third law is the main workhorse of ... 0 The torque that is applied to the pulley is created by friction between string and pulley. String has no mass but it has acceleration. So the force and torque that are applied to it must be zero. Free body diagram of string is shown below: If we write ∑τ about center of pulley, then we have: Consider that N=∫dN passes from center of pulley and its ... 0 Torque is force times the distance from the rotation point. (More specifically, it is the cross product.) The distance from the rotation point is just$r$, and the forces are$T_1$and$T_2$So, $$\tau_1 = T_1r$$ and $$\tau_2 = T_2r$$ 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$\theta = 0$. 0 In a cone-plate rheometer the plate is a disc, and the geometry means the strain rate$\dot{\gamma}$is constant everywhere. The stress is given by: $$\tau = \mu\dot{\gamma}$$ Since at equilibrium the viscosity is constant, that means both variables on the right of the equation are constant so the stress is constant everywhere on the plate. Now we just ... 0 A distributed normal force acting on a body due to contact with another body may be represented by a single normal force acting at the centre of action of the distributed force, but may just as validly be represented by a force-torque pair - i.e. by a force acting at a point on the surface, and a torque. In the latter situation, the force is a "normal ... 0 If the see-saw is hanging from the pivot point, at equilibrium we would have the following situation: The beam's overall length is$2L$and is attached to the pivot point$P$at the centre of the beam. I'll also assume the centre of gravity of the system is in$O$($mg$is the total weight of the system: beam plus$PO$) and that the distance between$O$... 0 In your calculation, torque and angular momentum are not particularly advantageous because the example is too simple - the force acts in the direction of the velocity. If we had gravity force, the concept of angular momentum and torque would be already quite useful, as it allows us to write down the equation of motion of the rotary motion without the need to ... 0 It's very easy: you need to consider torques if you cannon treat your objects as mass points. Here is a simplest situation: Imagine a body and two forces working on it, the forces being antiparallel and of equal strength, but not on the same line. The transaltional laws only tell you, that the center of mass will not accelerate, they cannot predict the ... 0 Torque is needed just as linear velocity is needed. Both indicate something happening at a distance. Torque is force at a distance, and velocity is rotation at a distance. In general, when you have one you also have the other. If you have a force through some axis, there is going to be a torque generated from that force away from this axis. If you have a ... 0 It looks like the pivot is slightly displaced from the beam's center of mass. If the center of mass of the beam is displaced slightly below the pivot by some distance$D$, then just pick the pivot as the center of mass and calculate the torque from that small displacement as the angle of the beam changes. The force$g\, M$of the center of mass is ... 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 ... 0 You have placed some strong restrictions in order to develop your question. If you remove those restrictions, there will be many situations in which torque calculations are necessary. Consider a static equilibrium situation with a uniform board of length$\ell\$, resting on two support points in a gravitational field. One support is at the left end of the ...

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