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Well of Death: Physics behind it
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[QUOTE="jack action, post: 6824502, member: 240508"] First, I'm not sure where you got that this equation is for ##v_{max}##; it is for ##v## in general. Second, the 'true' equation is: $$\frac{v^2}{rg} = \frac{\sin\theta + \frac{f}{N}\cos\theta}{\cos\theta - \frac{f}{N}\sin\theta}$$ Where ##f## is the friction pointing down the slope, which means a negative value is possible; it only means it is going against the slope. What is the value of ##\frac{f}{N}##? Well: $$mg = N\cos\theta - f\sin\theta$$ Therefore: $$\frac{f}{N} = \frac{\cos\theta - \frac{mg}{N}}{\sin\theta}$$ Note how it can have a negative value. Putting it in our original equation: $$\frac{v^2}{rg} = \frac{\sin\theta + \left(\frac{\cos\theta - \frac{mg}{N}}{\sin\theta}\right)\cos\theta}{\cos\theta - \left(\frac{\cos\theta - \frac{mg}{N}}{\sin\theta}\right)\sin\theta}$$ Or: $$\frac{m\frac{v^2}{r}}{N} = \sin\theta +\frac{\cos\theta - \frac{mg}{N}}{\tan\theta}$$ Isolating ##N## we get either: $$N= \frac{m\frac{v^2}{r}\tan\theta + mg}{\sin\theta \tan\theta +\cos\theta}$$ or: $$N = \frac{m\frac{v^2}{r} + \frac{mg}{\tan\theta}}{\sin\theta + \frac{\cos\theta}{\tan\theta}}$$ [LIST] [*]In the first form, assuming ##\theta=0##, we get ##N=mg##. [*]In the second form, assuming ##\theta = 90°##, we get ##N=m\frac{v^2}{r}##. [/LIST] That's it. There is no ##v_{max}##. Whether ##\theta=0## or ##\theta=90°## or anywhere in between, you can go as fast as you want in any case. Only the normal force will change, which is independent of ##v## when ##\theta=0## and independent of ##g## when ##\theta=90°##. But let's isolate ##\frac{f}{N}## in our original equation instead: $$\frac{f}{N} = \frac{\frac{v^2}{rg}\cos\theta - \sin\theta}{\frac{v^2}{rg}\sin\theta + \cos\theta}$$ Or: $$\frac{f}{N} = \frac{\frac{v^2}{rg} - \tan\theta}{\frac{v^2}{rg}\tan\theta + 1}$$ And we know that to remain static (our assumption in the first place i.e. no sliding): $$\mu_s > \left|\frac{f}{N}\right|$$ So: $$\mu_s > \left|\frac{\frac{v^2}{rg} - \tan\theta}{\frac{v^2}{rg}\tan\theta + 1}\right|$$ But that doesn't indicate a ##v_{min}## either, just a ##\mu_{s,\ min}##. That only means that the required friction coefficient will vary between ##\frac{v^2}{rg}## at ##\theta=0## (such that the vehicle doesn't slide out the curve) and ##\frac{rg}{v^2}## at ##\theta=90°## (such that the vehicle doesn't slide down, or "into" the curve). The only notable thing is that if ##\frac{v^2}{rg} = \tan\theta## then ##f=0## and therefore no friction is required.We can find a ##v_{min}## and ##v_{max}## such that the vehicle doesn't begin to slide in either way. Assuming a given ##\mu_s##, we go back to our original equation (converted to the ##\tan## equivalent): $$\frac{v_{min}^2}{rg} = \frac{\tan\theta + (-\mu_s)}{1 - (-\mu_s)\tan\theta}$$ $$\frac{v_{max}^2}{rg} = \frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}$$ In both cases, the right-hand side cannot be negative, so: $$\frac{v_{min}^2}{rg} = \frac{\max\{\tan\theta; \mu_s\} - \mu_s}{1 + \mu_s\tan\theta}$$ $$\frac{v_{max}^2}{rg} = \frac{\tan\theta + \mu_s}{1 - \mu_s\min\{\tan\theta;\frac{1}{\mu_s}\}}$$ I think I got it all right. [/QUOTE]
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