# Tag Info

91

You're talking about a device (in helicopters the tail fan imparting horizontal thrust) that counteracts the torque imparted on the main rotor (and therefore on the helicopter) by the surrounding air as the main rotor is dragged through the air. You propose instead to impart an opposite torque through a reaction wheel. That would indeed impart an opposite ...

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What luck! Just yesterday I was thinking about this exact same phenomenon whilst watching the film 'The Imitation Game'; the title sequence contained a moving tank. When I was little, I used to observe this all the time; not in wheels however, but in caterpillar tracks: Notice how, when a segment of the track touches the ground, it just stays there, in ...

39

There is another nice way of seeing this mathematically. It is not too hard to show that in the body frame, there are two conserved quantities: the square of the angular momentum vector $$L^2 = L_1^2 + L_2^2 + L_3^2$$ and the rotational kinetic energy, which works out to be $$T = \frac{1}{2}\left( \frac{L_1^2}{I_1} + \frac{L_2^2}{I_2} + \frac{L_3^2}{I_3}... 38 You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling. What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily ... 37 Buzz's answer is correct in that there's no such thing as a perfectly rigid body in relativity. But even more importantly for your question, a body in uniform motion does not feel any kind of squeezing force, even if it's moving very quickly. Consider two spaceships moving past each other at high speed. Ship A will see ship B compressed, and by the symmetry ... 29 It depends what you consider to be the "pivot" about which the rotational kinetic energy is calculated here: If you choose the pivot as the end of the rod that is physically held in place, then you only have to consider rotational kinetic energy, since if you think about what rotational kinetic energy represents, then it should be clear that the ... 23 The rectangular prism is a rigid body. The equations of motion of a rigid body around its center of mass are given by: (Please, see for example: Marsden and Ratiu , (page 6).$$I_1\dot\Omega_1=(I_2-I_3)\Omega_2\Omega_3I_2\dot\Omega_2=(I_3-I_1)\Omega_3\Omega_1I_3\dot\Omega_3=(I_1-I_2)\Omega_1\Omega_2Where \Omega_1,_2,_3 are the angular ... 23 You have to understand that concepts like rotational kinetic energy are just shortcuts to solve problems efficiently. In Classical Mechanics, we start by defining concepts like kinetic energy on point particles, with no dimension; when we need to extend this concepts to macroscopic objects, like the rod in your problem, the rigorous way to do it is to think ... 22 As Yashas Samaga said, it will not stop on a smooth, but frictional surface. It will stop however on an actual rough surface (as it does in reality – e.g. a steel marble rolling on a rough stone surface will come to a halt quite quickly, although drag / rolling friction is as low as on a smooth glass plate, where the marble would indeed roll very far). The ... 21 If the yoyo is spinning without winding the string, then it must be that the friction between the inner barrel and the string is insufficient to lead to winding. A flick of the wrist or some other movement can lead to an increase in the friction and the beginning of winding. Once winding begins frictional forces between the string and barrel increase ... 19 The instability inherent in the medium length axis or \prod_2  as shown above is discussed in detail in Marsden and Ratiu, which is where the image is from. The unstable homoclinic orbit that connect the two unstable points have intersting features. Not only are they interesting because of the chaotic solutions via the Poincare-Melnikov method that can ... 18 Having a non-zero relative velocity is fine as long as the distance between the points isn't changing. This certainly holds for a rotating rigid body. As another example, take a ball on a string and rotate it in a horizontal circle. Is the ball moving relative to you? Yes. Is it moving towards or away from you? No. Therefore this part If they have a non-... 17 Assumptions made in this answer: By rough surface, you meant a flat surface which has friction. The cylinder/sphere/disc/etc. are ideal; they do not deform. This is my reasonable guess; I am aware of the terminology used in Indian high school textbooks and exams (I am from India too) but you should still edit your question and make it clear. When a ... 17 Since the rod's center of mass is changing, does this mean that it also has translational kinetic energy? Yes. You have both translational kinetic energy and rotational kinetic energy. The translational kinetic energy is due to the translational motion of the COM. Although the rod is rotating about the hinge point, it is also rotating about its COM. It is ... 16 In a scenario of pure rolling of a rigid wheel on a flat plane, you don't need any friction. Once the wheel is rolling, it will continue to do so, even if the friction coefficient becomes zero. If this were not the case, you'd be violating conservation of angular momentum. There is no force, no torque and therefore no work done. The more interesting case is ... 16 Perfectly rigid bodies are not possible in relativity, although this is not directly related the Lorentz contraction mentioned in the question. One immediate consequences of relativity is that no signal can travel faster than the speed of light; and this actually rules out perfectly rigid bodies. The reason, although it may not be instantly obvious, is ... 15 The above answers are all good but i want to give another example which really helped me with understanding what does it mean that the point of contact has a velocity of zero. Think of the 'spinning circular object' not as a ball, but instead think of it as a star polygon with an infinite amount of edges, for the sake of the example a very large number will ... 14 Here's a straightforward but somewhat computational way. There are two steps. (1) Show how to define the angular velocity vector in terms of rotation matrices. (2) Write a general rotation in terms of Euler angles. (3) Combine (1) and (2) to get an expression for the angular velocity vector in terms of Euler angles. Step 1. Recall that if \mathbf x(t) is ... 14 There's an alternative to @MichaelSeifert method which uses angular momentum and moments of inertia: it is to deal with the vector \vec\omega directly as we are interested in the evolution of this vector. One can express the kinetic energy and length squared of \vec L as \begin{align} T&=\frac{1}{2}\left(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2\... 13 Keep in mind that the screw axis does not have to pass through the body. For your example place the axis of rotation parallel to the x axis straight above the cylinder, then rotate the cylinder 180° about it. The result will be equivalent to 180° rotation about the x axis followed by a translation along the z axis by twice the distance from the x axis to the ... 12 Proposition: Given an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body, and wrt. to an arbitrary choice of Cartesian coordinates x, y, and z), then the diagonal elements I_{xx}, I_{yy}, and I_{zz} of the inertia tensor satisfy the triangle inequality, I_{xx} +I_{yy} ~\geq~ I_{zz}, \qquad I_{yy} +I_{zz} ...

12

Consider a point $P$ on the surface of the wheel. If you look at the horizontal velocity of that point in the frame of reference of the wheel (axis stationary), then for a wheel of radius $r$ with angular velocity $\omega$ that point will have horizontal component of velocity $$v_h = r\omega\cos(\omega t)$$ The linear velocity of the wheel $v = \omega r$. ...

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The problem here is that no force can be perfectly impulsive because then the distance moved while the force was being applied would be zero and therefore the work done would be zero. If we assume some finite impulse $I$ then: $$I = F t$$ where we are applying the force $F$ for a time $t$. The perfectly impulsive force would be the limit $t \to 0$, and ...

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Its really a great question and let us go by simple definition. Defination: As per what wikepedia says: A yo-yo (also spelled yoyo) is a toy which in its simplest form is an object consisting of an axle connected to two disks, and a length of string looped around the axle, similar to a slender spool. Working: But what really makes it work? In simple it is ...

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If the velocity of point B relative to point A is always at right angles to the line AB joining them, then the distance does not change.

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you wrote: "How can [the velocity of the contact point] be zero when it's in continuous motion?". However, you should keep in mind that motion is relative and therefore your question should be actually read as: "How can the velocity of the contact point be zero relative to the contact surface, when it's in continuous motion relative to its axis of rotation?"...

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You are making a major flaw here. The friction between the 2 blocks is not going to be $10N$. It is going to be something, but we will have to calculate it. Assume the friction force to be $f$ such that the acceleration for both the blocks is same. Now, equations for $m_2$ and $m_1$ separately are: $$F-f = m_2*a = 3a$$ $$f = m_1*a = 5a$$ Solving both ...

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As far as the laws of physics are concerned, you could do it if every so often you use the stored angular momentum in the flywheel to quickly reverse the direction of the main rotor and then start building up angular momentum in the other direction. Disadvantages: Oh boy, where to start? You need symmetric, and thus probably less efficient, rotor blades, ...

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