# Tag Info

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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 ...

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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 ...

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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 ...

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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 ...

3

The answer to that is because the moment of inertia is not the same for the solid cylinder than for the hollow one. As you write the formula for the moment of inertia, it depends on the distribution of the mass. The further away the mass is from the rotation axis, the more contributes to the moment of inertia (as in distance squared $r^2$). So, since the ...

3

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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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 ...

3

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 ...

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 ...

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 ...

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 ...

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Mass 1 is given an initial velocity $v_1(0)=v_o$. You want the velocity $v_2(t)$ of mass 2. One way to do this is to break up the motion of mass 2 into the motion of the center of mass of the two-mass system, and the motion of mass 2 relative to the center of mass: ...

2

The centrifugal force on the ring is the pseudo force when in the ring's reference frame, which causes it to move outwards, given by $$\vec{F} = m\frac{v^2}{r} = mr\omega^2$$ Where m is the mass of the object, v is the tangential velocity of the object, and omega is the angular velocity To find the time required for the ring to fall off, you need ...

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 ...

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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. ...

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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} ... 1 In the lab frame there is no centrifugal force. The ring goes outside because it lacks of a centripetal force. Let's take a step back: if the rod spins slowly then the ring does not slide out because of the friction. In this case the friction is a centripetal force, this means that it is responsible for keeping the ring on a circular motion. If you increase ... 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 = ... 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 ... 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). ... 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 ... 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 

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