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Please explain why 𝑟 is a function of 𝜃 Unfortunately, there is not much content to the statement. Basically, both $\theta$ and $r$ are functions of $t$. And if $\theta$ is an invertable function then $t=\theta^{-1}(\theta(t))$ so $r=r(t)=r(\theta^{-1}(\theta(t)))$. Now, we can write $R=(r \circ \theta^{-1})$ so then $r=R(\theta)$ shows that $r$ is a ...


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You've got $r^2\dot{\theta}$ constant. Unless that constant is zero, this means that $\dot{\theta}$ never changes sign, which means that $\theta$ is either always increasing or always decreasing as a function of $t$. Therefore you never have to worry about the possibility that $\theta(t_0)=\theta(t_1)$ with $t_0\neq t_1$. The only exception is where the ...


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Typically you have three variables, $r, \theta, t$. You are correct that r cannot be a function of $\theta $ if you limit the range of $\theta$ to for example $[0,2\pi]$. If you do not project to this interval then you can take $t$ as the independent variable and have $r$ depend on $t$ via $\theta $. This is a fairly standard approach to derive Kepler orbits....


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The two equations are mutually contradictory, so they cannot both be correct. Your teacher was wrong to say that 𝑁cos𝜃=𝑚𝑔. EDIT in response to Aaron's excellent correction. Sorry, I should have said that the two equations are mutually contradictory so they cannot both be correct at the same time. You are confusing two situations (or I was!), one in ...


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This is one of the many confusions that new students to physics face, as many of the first examples of objects on inclines do involve $N=mg\cos\theta$, so students get used to using that and forget why we can even say this in the first place. Newton's second law is a vector equation that says the net force acting on an object is proportional to the ...


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I would try to give a rather intuitive way to approach the question. Say the object is rotating in a circle with a tangential acceleration $a_t$. Here $a_t$ acts to change the speed of a particle in the tangential direction (say) changing it from $v_0$ to $v$. Now say at that instant the tangential acceleration ceases to exist then what can we say about ...


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No, the spinning of the ring itself does not suddenly create a force on the astronaut. It is the tangential velocity of the astronaut that makes him hit the ring, as he can not continue traveling straight due to the ring being in front of him, which ultimately leads to a change in direction and a corresponding force to make that change. In short, astronaut ...


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Centripetal force acts towards the centre and necessary to keep an object in circular motion. Correct. But centrifugal force is equal to the magnitude of centripetal force and acts in outwards direction and hence counteracts the centripetal force. The centrifugal force is an apparent, or fictitious force, that seems to be responsible for the ...


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Centripetal force is the force that is applied by the constraint that is responsible for circular motion. If the "constraint" somehow becomes inefficient at its job at a point (like, if the rope attached to the stone snaps, or a car undergoing a turn skids or there isn't enough friction) the object will continue to move in a straight line, namely the ...


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Centripetal force is necessary for an object to stay in circular motion because the only thing that the centripetal force is responsible for changing is the direction of the velocity, and not the magnitude. The centripetal force vector points perpendicular to the velocity vector at every instant in the object's circular path. The centrifugal force isn't an ...


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Yes. There is a maximum acceleration that your tires can provide.Be careful with sharp turns at high speeds where the centripetal acceleration is close to this limit. Any additional acceleration (or breaking) may cause the vechicle to slip out of the turn. If you decrease your speed, centripetal acceleration decreases and you will be more safely within ...


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It has to do with this: In a perfect world, turns would all be banked so you could traverse each at full speed and not skid sideways off the road. Since it is not always possible to bank the roadway surface at the right angle to support a full-speed turn, the reduced speed limit in a turn is set to accommodate the existing bank angle and turn radius minus a ...


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It's hard to say whether or not what you talk about is "due to inertia". Inertia is essentially just mass, but I think the term is usually either misused or just unclear when it comes up, so I try to stay away from it. We we can say is that if a ball is floating in space and rotating (as viewed from a non-accelerating reference frame), then it will continue ...


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A rigid body rotating about its center of mass has angular momentum. You need torque to change this angular momentum just as you need force to change linear momentum. These statements say nothing about mass or inertia on their own $$ \begin{aligned} \boldsymbol{F} & = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{p} \\ \boldsymbol{\tau} & = \tfrac{{\rm d}}{{...


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