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You got the signs wrong - the centripetal acceleration is towards the center of motion and enters the equation from the other side. This is a common mistake in analyzing free-body diagrams - you must include only the forces that act on the object. For every force you include in the free-body diagram you must be able to answer the following question: What ...


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In the absence of friction there are only two forces acting on the car - its weight $mg$ and the normal force $N$. If the road is at an angle $\theta$ to the horizontal then the horizontal force on the car is $N \sin \theta$. If the car is travelling at speed $v$ around a circle of horizontal radius $r$ then we must have $$\displaystyle N \sin \theta = \frac ...


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On a banked road, the normal force from the road has two components. On a friction-less road the vertical component supports the weight of the car. There is only one speed at which the horizontal component will supply the required centripetal force. You can't change one component without changing the other.


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Sure, you can always convert between centrifugal force in a rotating frame of reference and centripetal force in a non-rotating frame. You just have to consider all the acceleration effects. Not sure I followed your explanation but it seems to be missing the tension due to the inward acceleration. Here's how I would think of it: there's a drooping helicopter ...


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Indeed, the normal force is changing with orbital speed. As the object is flung more outwards with higher speed in the same circular path, it applies a larger force on the banked surface. Meaning, the banked surface has a larger load to hold back against in order to avoid breaking. The normal force is the name given to the forces that come into existence in ...


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Look at this free body diagram, from here you obtain $$\tan \theta=\frac{m\omega^2r}{mg}$$ Thus if the angular velocity $\omega$ increases the angele $\theta$ increases


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In order to understand from where the equation for acceleration comes from, first draw the free-body diagram of a seat. In the diagram below, $\theta$ is the displacement angle, $G = mg$ is the body weight, and $T$ is the tension (pull) excerted by the rope. In equilibrium, the seat does not move in vertical direction. First Newton's law for vertical ...


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Your first equation contains no torques and is not true. For torques about a point between his feet: $({N_2} – {N_1})(d/2)$ = dL/dt = (d/dt)(mvh) = mh($v^2$)/R where I'm taking (L) as the angular momentum about the reference point and h as the height of the center of mass. (This is consistent with the book solution.)


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In general, the angular momentum of a system of particles about a reference point is equal to the angular momentum of the center of mass (CM) about the point plus the angular momentum of the motion about the CM. Using the CM as the reference point, the angular momentum of the CM is always zero so the angular momentum of the system is the angular momentum ...


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This is just food for thought important enough to share as an answer, rather than a comment: World renown astrophysicist Kip Thorne discuses how a rotating universe would be the equivalent of a rotating Earth. During the Gravity Probe-B experiment. https://www.youtube.com/watch?v=Wvnf6BUsyyo


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The problem you are running into is the word "outward". Outward is not a fixed direction in space, it is the radial basis vector in polar coordinates which changes from point to point. So if you write your physics equations in terms of "outward" then you are using polar coordinates rather than standard Cartesian coordinates. The ...


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There are (only) four forces in nature. Gravitational force Electromagnetic force Weak force Strong nuclear force If you cannot explain the origin of a force from this list then it is not a real force. This is that simple.


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In the case of the bead on the rod the force that causes it to move outwards is the reaction force of the rod acting on the bead that causes it to move in the way it does (I mean that is the only net force component here, so it has to be this). We can see this by looking at Newton's second law in polar coordinates: $$\mathbf F=m\mathbf a=m\left(\ddot r-r\dot\...


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(1): If the cable that links the rotating object to the center is broken, the object moves in a straight line. In reality, if the cable is broken before the movement starts, there is no rotation at all, only straight movements. There is some tangential force to initiate the movement from the rest. And a centripetal force along the cables to avoid the ...


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In this answer I will proceed as follows: I will discuss inertia in terms of everyday life experience. After that I will address the specific case mentioned in the question: ball on a merry-go-round, with a string. The human psychology has the following quirk. When something is everywhere, you are likely to end up unaware of it. It becomes so internalized ...


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You are confusing what is real and what is imaginary and as shown in your statement From the merry go round, it is easy to explain the motion of the ball, and say it is the centrifugal force that makes the ball move outwards. There are forces acting on bodies which are the same irrespective of the reference frame, constant velocity, accelerating, rotating ...


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In the absence of any force (external observer) the object will simply remain on its (straight, constant velocity) trajectory. To the co-rotating observer this looks as if the object is "moving away" from him, but seen from the outside this is simply an object with a non-zero velocity relative to the merry-go-round moving happily along in a ...


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