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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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Typically there are two kinds of transformation that do not change the outcome of situation. Think of a force vector $\vec{F}$ passing through a point $\vec{r}_A$. Any translation along the line of the force, in the direction $\vec{e} = \frac{\vec{F}}{\| \vec{F} \|}$ will not change the outcome. Any rotation about the line of the force would also not ...


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First thing you need to do is convert the spherical coordinates into cartesian and then derive a 3×3 rotation matrix from the axis angle information. You can manipulate the rotation matrix using elementary rotations and then get the axis angle for the final orientation. Then you convert the axis angle to spherical coordinates for use in your system. Here ...


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In general it changes although the reason is not exactly because its projections changes. For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change ...


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Direction of a vector is determined by the components themselves. Now if the components are changed the direction gets changed by the above definition. All this is with respect to one reference frame.


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The frictional force is given by $f = \frac 1 3 mg \sin \theta$ when the no slipping condition ($a=r \alpha$) is satisfied. If the component of weight down the slope, $mg \sin \theta$, was equal and opposite to the frictional force then there would no net force on the cylinder and so the centre of mass of the cylinder would not be accelerating ($a=0$). ...


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The above answers do not take into account the vector nature of $\omega$. I venture to provide an answer that hopefully should satisfy the questioner. We Consider a point at colatitude $\lambda$ on the Earth's surface in the Northern hemisphere. We draw a local coordinate system at this point such that the x-axis points south, the y-axis points east, and the ...


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First of all: your solution is correct, the friction is one third of what it would have been for a "non-round", as you call it, body. How does friction know. I like such questions, but I don't see a convincing simple reasoning right now (but there probably is one), so I will use a little of computation. I hope it's different (less computating^^) than your ...


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Friction acts to oppose the two surfaces in contact from sliding past each other. It will be no greater than necessary to do that.


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The only thing I've figured out, is that the object will rotate but not moving, if the center of mass is on the line defined by the $A$ point, and the normal vector of $\vec{F⃗}$. This is not necessarily true. Look at the free body diagram. Decompose $F$ into $x$ and $y$ components. Newton now tells us that: $$ma_x=\Sigma F_x$$ $$ma_y=\Sigma F_y$$ ...


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The laws describing the movement of the body are: $$\sum \vec F=m\vec a_{cm},$$ and $$\sum \vec \tau= \frac{d\vec L}{dt}$$ wehre $\vec F$ is the external force, $\vec \tau$ the external torque, $\vec L$ the angular momentum and $\vec a_{cm}$ is the acceleration of the the center of mass. As you can see from the first formula, the center of mass will always ...


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The moment of Inertia $I$ is defined such that the rotational energy of the object about it's pivotal axis is $E = I \omega^2$ where $\omega$ is the rotational speed. In the case of a small object moving in a circle of radius $r$, the rotational energy is given by $$ E = \frac{1}{2} m v^2$$ and using the relation $v = r\omega$, we have: $$E = ...


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I assumed that the angular velocity (and so $α$) is the same it I take as pivot the center of mass or the point $O$. If this is the case than parallel axis theorem can be used You are mixing rotation and circular translation. Angular velocity is defined with respect to the (instantaneous) axis of rotation, which you cannot choose at will. It is ...


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The equation of motion $$ \text{torque about stationary geometrical point O} = \text{moment of inertia w.r.t. O} \times \text{angular acceleration w.r.t. O} $$ is valid only if the motion of the body is planar rotation around an axis that passes through O. This is the case if the point O is taken to be point of contact of the body when rolling without ...


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As an unbalanced force, $\mathbf{f}$ acts to accelerate the disk. Since it is located at the bottom of the disk, O must accelerate as well and is therefore in a non-inertial frame of reference. That non-inertial frame will have a fictitious forces appear that oppose acceleration. We can draw a force $\mathbf{f'}$ that acts through the center of mass in ...


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As friction takes place, it implies that the center of mass is decelerating. To simply apply the Newton's law, one has to stay in the inertial frame. Alternatively, if one discusses the problem in a non-inertial frame of reference, some inertial force $\vec{f}_{int}$ should be properly added to the analysis to restore the law's validity. In the following, ...


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When doing this sort of problem you can add two forces acting at the centre of mass whose resultant is zero. This system of three forces can now be viewed in the following way. The frictional force $f$ is exactly equivalent to a force of the same magnitude whose line of action passes through the centre of mass of the disc (shown in blue) and a pair forces ...


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Rolling of a circular body, on a flat surface, and without sliding results in $v_\text{tan}=v_\text{cm}$ where $v_\text{tan}=ωr$ is the tangential speed of any point on the rim of the body in the center-of-mass frame of reference. This is understood by studying the motion in the c.m. frame: there, the flat surface has velocity $v_\text{cm}$ (backwards). The ...


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Are you confused about how you get into this? $$\begin{align} v & = \omega\,r \\ ({\rm m/s}) & = ({\rm rad/s})\,({\rm m}) = ({\rm m\,rad/s}) \end{align} $$ Radians are not units with dimensions. They can be seen as $({\rm rad}) = ({\rm m/m})$ like arc length to radius. This makes to above right hand side equivalent to the left hand side.


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This is all true because the ratio of a circle's diameter to circumference is constant (pi). The unit of radian is actually chosen so that l = r.theta. So if you go theta radians around a circle you travel r.theta. For the angular to linear velocities, think of a disc rotating at an angular speed omaga. then the further out from the centre of the disc you ...


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You can look at this situation in two ways. First in terms of absolute motions $\Omega$ and $\omega$, and secondly it terms of the relative motion $\dot{q} = \omega - \Omega$. The results are the same The cylinder is spinning with $\omega$ speed and translating with $\Omega R$ speed. The combined momentum and kinetic energy is Linear Momentum: $$p = m R ...


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Yes.. The two angular velocities are independent of each other... It's basically like earth rotating around the Sun... Earth's speed of revolution around the Sun is independent of its speed of rotation about it's own axis.... I'm not very sure about your first equation where you add the two angular velocities... As Leverl said, the two motions are ...


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It has already been somewhat answered in this other question of yours: Parallel axis theorem and Koenig theorem for angular momentum (see also my comment to the answer). Angular velocity, like translational velocity, has to be defined by reference to a frame. In your question, you state the angular velocity about the principal axis is $ω$. So its angular ...


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It should be easy to see that if it "disintegrates" into dust particles, it is no different than if you have a rotating object, held by a string, and the string is cut. We know that the angular velocity (momentum) $\omega$ will be converted into a tangential velocity $v$ and the particles will have a momentum $mv$, in addition to the momentum generated by ...


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Let's just say, for the sake of argument, that the equator has a linear eastern velocity of 1,000 mph (actually 1,039). If you release a ball (presumably fired from a cannon) southward from a latitude of 70° north, your ball has an eastern velocity of about 342 mph (cosine of 70° = 0.342). As your ball travels over 45° north, the ground beneath it will have ...


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The direction of deflection of the ball should be independent where the observer is. Left or right deflection is with respect to an observer facing in the direction the ball is moving. Are you asking about a ball being thrown from the northern to the southern hemisphere? In that case the deflection would change from deflecting right to deflecting left as ...


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When youare talking about angular momentum about any axis(inluding one that passes through center of mass), you carry out the "formula":$\vec{L}=\vec{L_{com}}+I_{com}\vec{\omega}$ where direction of $\vec{L_{com}}$ and $I_{com}\vec{\omega}$ is to be kept in mind, during vector summation.While we calculate $\vec{L}$, about COM, $\vec{L_{com}}$ becomes zero ...


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When you talk about a rigid body the distance between any two points is fixed. That implies angular velocity w is same for all points. Now ony moment of inertia changes: I=Icm+md^2 [d is distance between cm and the point] This is the parallel axis theorem. Get I and angular momentum=Iw


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The angular momentum will be different: however, you will be able to calculate it with the parallel axis theorem. Measure the distance from your new axis to the center of mass and call it $d$. Your new rotational inertia $I$ can be calculated from the rotational inertia around the center of mass $I_{\mathrm{cm}}$ using the formula: $$I = I_{\mathrm{cm}} + ...


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The vector $\mathbf r$ does change, even though its magnitude is nearly constant. Most importantly, the component of $\mathbf r$ which is perpendicular to the rotation axis is decreasing in this example. This fact leads to the explanation for which you are searching. Another way of seeing this would be to use this definition of angular momentum, which ...


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The tension force at any point in the rod provides the force that keeps the parts of the rod further from the centre of rotation moving in a circle. For points closer to the centre there is more mass to swing around i.e. a larger centripetal force is required, so the tension is larger. I'm assuming your thinking of a rod rotating in a horizontal plane. In ...


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As you mentioned, $\vec {L_o}_⊥$ is proportional to $\omega$. So the middle term means $\vec {L_o}_⊥$ varies by varying angular velocity. You can also derive this. formula like this: $$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega} \frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } ...


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Well, my reasoning may be a bit circular, but I'd say that for the linear and angular velocity not to be independent, the one needs to be a function of the other: $v_{cm}=v_{cm}(\Omega)$. I can only think of cases where the relation would be directly proportional. Two examples: A ball or cylinder rolling over a surface without slipping: in this case ...


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I think that I understand what you are asking and I think that it is a good question. If the centre of mass of the disc and the point mass moves a distance $x$ then the work done by the force is $Fx$ in both cases and yet the disc has gained some extra kinetic energy because it is rotating. If the centre of mass of the disc moves a distance $x$ the point ...


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There's one thing in you question that doesn't quite add up and that is when you say exert a horizontal force at a certain distance r from the center of mass. If you are pushing a disk anywhere you push it (on the curved surface at least) is going to be a distance r from the centre of mass because the whole surface is a distance r from the centre of mass. ...


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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 ...


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Everything you have derived is correct. The reason for your perceived paradox is, I believe, a confusion between force and power. The same force can produce more power if it is being exerted at a greater velocity. When you exert a force at a radius r from the CM, the point of application of the force will accelerate more quickly than the CM, allowing the ...


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I don't quite understand what you are asking. In your diagram you have $$ \begin{aligned} a_{tangential} & = \dot{\omega} \ell \\ a_{centripetal} & = \omega^2 \ell -g \end{aligned} $$ what else to do want to know? If the angle is changing (and therefore your acceleration are not aligned with X and Y) then use $$ \begin{aligned} a_X & = ...


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The term gyroscope is more often associated with sensors that can sense angular velocities, and these can be used in a control feedback system with actuators such as reaction wheels or control moment gyros that provide the application of torque to the body of the spacecraft and thus control attitude. Thrusters indeed can provide translational force ...


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Let's look at the hodograph of a constant radius & constant velocity motion. Left: trajectory of one of the masses. Right: hodograph, i.e. locus of the velocity vectors. Now, let's look closer at how the velocity changes during a small time interval $\mathrm dt$. A force is needed to rotate it (difference between the brown and red arrows). If you ...


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A body can have parallel velocity on the instantaneous axis of rotation. This parallel velocity is sometimes designated with a scalar pitch value $h$, such that $\vec{v}_O = h \vec{\omega}$ Consider a point O on the IAR and a point P outside of it. You have $$ \vec{v}_P = \vec{v}_O + (\vec{r}_P-\vec{r}_O) \times \vec{\omega} $$ Now I can prove that given ...


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The centrifugal force is given by $mR\Omega^2$. Find the value of $R$ for which this force is greater than the maximum frictional force possible.



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