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In the frame of reference of the body, is the centripetal force felt or is only the centrifugal force felt? It depends on what you mean exactly. Consider, for example, the amusement park ride Dumbo at Disneyland: . On this ride, passengers sit in mini Dumbo replicas and are swung around in a circle. What forces do they feel? Well, firstly, they ...

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In the frame of reference of the body, is the centripetal force felt or is only the centrifugal force felt? In the frame of reference of body total force acting on it is $0$ because it is at rest w.r.t itself. It means net force acting on it should be $0$ but how that would be possible if only centripetal force is acting, how would Newton's law hold. ...

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Here's a motivation for where the inertia tensor $I=(I_{ij})$ (and by extension moments of inertia) comes from. It's a quantity that's analogous to mass for rotational motion in the sense that the kinetic energy of a rotating object is essentially proportional to the inertia tensor times the square of the body's angular velocity. More precisely ...

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I think you are confussed between mass moment of inertia and area moment of inertia. The first is an equivalent of mass in angular direction and is defined as $\int_V{r^2\rho dV}$. An angular equivalent of $F=ma$ is: $$\tau=I\alpha$$ where $\tau$ is torque (angular equivalent of force, with units $[Nm]$), $I$ is mass moment of inertia (angular equivalent of ...

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The angular momentum of a single particle with mass $m$ in motion about an axis, with angular speed $\omega$, a distance $r$ from the axis, is $L = r (m v) = m r^2 \omega$. When we consider an extended body, the sum up the contribution ($m r^2$) from each particle in motion inside the body, and this is the moment of inertia. More generally, \begin{align} ... 0 The contact point of the disk with the plane has null instantaneous velocity This implies that there is no slippage, and as such there are no non-conservative forces doing work on the disk. Assuming the disk is perfectly rigid and is not being subjected to any linear or angular accelerations, the disk will continue to roll forever, and will not come to ... 0 Carl's first paragraph answers your question (though I disagree with his second paragraph) so this is just an addendum to Carl's answer. It sounds to me as if you are describing an ideal conical pendulum. You're correct that no work is done because the two forces, the string and gravity, act at right angles to the direction of motion so \vec{F}.\vec{dr} ... 0 There is force exerted (along the string) thru the point at which the mass's string is attached to whatever is holding it up. If there weren't, yes the ball would fly off. Gravity is a separate force, and will cause the mass to oscillate about the vertical axis. So in fact you won't be able to achive uniform circular motion. Take a look at the toys you ... 0 A rolling disk will come to a stop eventually because any incidental friction will decellerate the center of mass. Ideally with a flat surface, and constant motion there should be no change as there will be not friction required to keep the disk rolling. In real life though, for sure a rolling disk will stop eventually. 1 If \vec{p} the vector connecting the center of mass of b1 to the center of mass of b2 then you must have \vec{v}_2 = \vec{v}_1 + \vec{\omega}_1 \times \vec{p} \\ \vec{\omega}_2 = \vec{\omega}_1  \vec{a}_2 = \vec{a}_1 + \vec{\alpha}_1 \times \vec{p} + \vec{\omega}_1 \times \vec{\omega}_1 \times \vec{p} \\ \vec{\alpha}_2 = \vec{\alpha}_1 $$0 There is no easy way to model a spinning coin and calculate these observations. It slows down mostly because of air resistance and friction(here you must take velocity dependent friction-angular velocity in your case-) and it moves due to the combination of torque of gravity(a.k.a. precession) and friction. Velocity dependent frictions generally gives you ... 0 from the point of any point on the rod , the rod is rotating about that point.refer to landau mechanics . 1 Let's say you roll a ball (of mass m) down an inclined plane of angle of inclination \theta and coefficient of static friction \mu_{static}. Then you know a force parallel to the inclined plane acts on the ball through its center of mass. Another force parallel to the surface acts in the opposite direction of motion as follows, The force \vec F = ... 0 Yes the static friction of the body is zero if it is rolling. This is because, the velocity of the lowermost point (i.e. the point of contact is zero by definition of pure rolling). So, as there is no tentative motion of the lowermost point, the static friction is 0. But whenever a force is applied, the lowermost point has a tendency to slide. As a result ... 0 When one talks about static friction one generally thinks of the coefficient that enters in the static friction force (\mu_{static}). The force that you can apply to a body without moving it is proportional to this coefficient ||{\vec{F}}||\propto \mu_{static}. You should not confuse this friction with the friction that occurs for a body that moves ... 1 Both approaches are equally correct in this case. F = mv^2/R  is just a consequence of the law for rotational motion, which says  \tau = I\alpha (Torque = Moment of Inertia * Angular acceleration). The former formula may be used in case the objects in consideration are point masses. But the latter, more general version of the formula is applicable for ... 2 When a disk or other object is rotating on a horizontal surface with constant velocity, there is no static frictional force. Your logic is correct: if there were a horizontal force, the center of mass would be accelerating. If the rolling object suddenly encounters a frictionless surface, it would continue to satisfy the rotating without slipping condition. ... 4 If the ladder is slipping on the floor as well as the wall, then the point of rotation is where the two normal forces intersect. This comes from the fact that reaction forces must pass through the instant center of motion, or they would do work. In the diagram below forces are red and velocities blue. If the ladder rotated by any other point other than S ... 0 If you wanted to measure the arc length swept out by a point on the rim of a wheel for a set amount of time, you would be correct; the faster one would cover more length in the same amount of time. It's unclear to me, but this may be the reason you believe the distances should be different. However, the "arc length of a cycloid" S=8r is calculated for the ... 3 The ladder falls because it experiences unequal moments from the normal reactions at both its ends. That is to say that the surface pushes on the ladder from the bottom as well as the side. In the absence of tangential contact forces such as friction, the ladder rotates and falls. To solve a problem with such a situation, you may choose any point as the ... 1 This is a case of non-stationary axis rotation. The axis is generally taken as the COM of the ladder as about this axis, the mass distribution is equal on both sides. 0 If you know the motion of a point A on a rigid body, with linear velocity \vec{v}_A and angular velocity \vec{\omega} then the formulas below will give you the linear momentum of the rigid body, and the angular momentum about point A. If the body is pivoting about A then \vec{v}_A=0, otherwise in the general case \vec{v}_A \neq 0 . Linear momentum ... 0 No. Centripetal force acts only when something is forced to move in a circle. In your case the rings are free to move radially out as you expect. Consider a coordinate system, with x axis along the rod of length \ell and y in the direction of motion of the end of the rod. If you count the distance from the center of rotation as r=\frac{\ell}{2} and the ... 4 The moment of inertia tensor is not constant in the external reference frame (http://en.wikipedia.org/wiki/Precession#Torque-free ) 1 Great question; I remember being so confused by this when I first took analytic mechanics. The components of the angular velocity "in the body frame" aren't zero because when one writes these components, one isn't referring to measurements of the motions of the particles in the body frame (because, of course, the particles are stationary in this frame). ... 3 There's another way to do this also, more akin to how spacecraft actually do it: Take a weight on a string, hold it up and spin it. You'll turn in the opposite direction. When you stop it you also stop turning. Of course this will produce an off-axis force that will be a real pain to deal with. Real spacecraft do it by means of a set of internal wheels ... 19 For those that are cat-challenged, here's an alternative explanation and demonstration you can try at home! This demonstration was taught to me by my math lecturer. All you will need is: A swivel chair and a heavy object (e.g. a big textbook) Stand on the seat of the chair (watch your balance now) holding the heavy object. Extend your arms forward ... 43 The astronaut can change his or her orientation in the same way that a cat does so whilst falling through the air. After the transformation, the astronaut is still and angular momentum is conserved. There is a rather beautiful way of understanding this rotation as an anholonomy i.e. a nontrivial transformation wrought by the parallel transport of the cat's ... 3 Other answers have pointed out other ways that might be more efficient, but one very simple way to do it is as follows: start with both arms parallel to the body. Then swing them both backward, up over the head, and then back down in front of the body, leaving them back in the starting position. After this manoeuvre, the body will be oriented in a slightly ... 3 Well, the angular momentum of a rigid body is equal to the sum of the angular momentum of the body around it's center of mass, plus the angular momentum of the center of mass. Having said that, suppose the rod is rotating about one end (I imagine a pendulum motion; correct me if I'm wrong), you can calculate the angular momentum by L = I \omega if you ... 0 Pebble will not move according to a inertial frame outside the disk which is rest with respect to ground,Since looking from this inertial frame,there is no horizontal force acting on pebble because of frictionless.Hence accoring to newton's 2^{nd} law pebble will stay in its state according to a observer from ground. But for an observer rotating along ... 1 I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass. To my understanding, your question is flawed. If a single force is applied to a rigid body under the ... 2 In that case the pebble won't move. If there is no friction, there won't be any forces between the pebble and the disk. 3 Let's assume that this whole setup is being viewed from an inertial frame and that if there is gravity, then it points perpendicular to the plane of the disk, then The disk will slide under the pebble, and the pebble will stay where it is. Why? Well in an inertial frame, Newton's second law holds. Since the force on the pebble tangent to the surface of ... 0 Explaining conservation of angular momentum is a good idea. I would like to add another explanation that is probably at the 9-year-old level. Imagine the top from, well... the top. Let's say it is going clockwise. Suppose the mass begins to tip to the right. In a short amount of time, the rightmost part of the top would have experienced a downward ... -2 A small shutting down on increasing resistance because the small generator has no ability to produce such current to perform such load. If the load is equal to the ability of the generator then it will not shut off. -1 Along time ago (1960's) I saw on TV a show where a scientist was explaining the working of a spinning top in front of an audience of children. He had a roundabout with a seat in the middle on which he sat a young boy. The boy held a metal bar with a wheel on its end (Quite heavy!) The scientist spun the wheel and rotated the roundabout. I cant remember ... 2 why we always choose the center of gravity of the bicycle be the rotational center. We do not do that always, sometimes it is better to use the point in contact with the ground or some other point. We use center of mass when it leads to simpler equations than the other points. In problems dealing with torques or rotations we use the theorem T: the sum ... 1 As is well-known from Newton's shell theorem, the gravitational field g(r)=\frac{GM}{r^2} outside a spherically symmetric mass-distribution is the same as if the total mass M sat in the center. It seems that OP wants to calculate the oblateness of Earth under the simplifying assumption that the backreaction (which the re-distributed mass has on Earth's ... 0 Download the file at https://www.dropbox.com/s/5l3lvqbal1ch2gm/Asymmetric%20Top.nb and run in Mathematica. Executing the command en[3,0], for example, yields the energies and vector representations of the J=3,m=0 wavefunctions in the basis$$\{|3,-3,0\rangle,|3,-2,0\rangle,|3,-1,0\rangle,|3,0,0\rangle,|3,1,0\rangle,|3,2,0\rangle,|3,3,0\rangle\}$$where ... 1 Assuming a mass moment of inertia of I_1 for the main rotor and I_2 for the secondary rotor, and a coefficient of drag of \beta_1 and \beta_2 respectively then the torque on the rotor shafts are$$ T_1 = I_1 \dot \Omega + \beta_1 \Omega^2 \\ T_2 = I_2 (\gamma \dot \Omega) + \beta_2 (\gamma \Omega)^2  where $\Omega$ is the main rotor speed, and ...

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The interchangeability of rotors for tourque might be nearly impossible because of the ma in f=ma.The rotor speeds would have to continuously vary ideally.Now if it were just a question of lift it would be in aronautical engineering. But then again as they say ,nothing is really impossible,you just have to find a way of doing it.

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The separation between the rotors does not actually matter. What matters is that the torque exerted by each of the motors on the respective rotor be the same and in opposite directions. Those torques then add vectorially and cancel out to give zero net torque on the helicopter. It's not clear to me exactly what you mean by "small". It is indeed possible to ...

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