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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 ...

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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 both centripetal and centrifugal forces are felt. does a body only feel the effect of pseudo forces in an accelerated reference frame? No the body feels all Pseudo forces and real forces in an ...

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I believe your confusion comes from a misunderstanding of the designation of a force as "centripetal". Any calculation of centripetal force is telling you how much force is needed to make a circular motion take place. This doesn't create the force. There is no guarantee that a force of the calculated size and direction actually exists! You need to go ...

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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 ...

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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.

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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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The Earth orbital speed could be increased by bringing the Earth closer to the Sun. That would require orbital maneuvering of the whole planet (planetary retrograde burn to go into a transfer orbit and another retrograde burn to circularize the orbit) which would be felt by the Earth population as acceleration. After the final orbit would be achieved, the ...

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The faster the earths orbit, the further it would go form the sun until it reached "escape velocity". The speed itself would make no difference to you anything you can feel. However, the distance from the sun, you would feel. The sun rotates around our galaxy core at one tremendous speed (and us with it), and you don't feel that.

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It's not missing, it's in the $\ddot{R}(t)$ matrix. It doesn't show up on its own when you do the calculation with matrices instead just vectors however. Vector equations The first equation should be written $$\mathbf{x}_A(t) = \mathbf{\phi}(t) \times \mathbf{x}_B(t)$$ (just reverse where you have $A$ and $B$) since given the position in frame $B$ you ...

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Partly because the magnitude of the gravitational force decreases as $\frac{1}{r^2}$, so as the distance from the center of the earth, $r$, increases, the magnitude decreases. The bigger reason for spacecraft is because they are constantly in free fall, and there is no way to feel gravity when you are falling freely. The spacecraft are falling and moving ...

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