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Well of Death: Physics behind it
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[QUOTE="NTesla, post: 6824361, member: 682219"] [B]Homework Statement:[/B] What will be the maximum velocity for a bike/car to go around in a well of death, when the wall of well of death is vertical, i.e. at 90 degrees ? [B]Relevant Equations:[/B] $$v_{max} = \sqrt{\frac{rg(sin\theta +\mu cos\theta)}{(cos\theta -\mu sin\theta)}}$$ While studying motion of car on banked curve, I was wondering, what will be the vmax when theta is equal to 90 degrees or is close to 90 degrees as it happens in a well of death which is organised in a village fair. On a banked road with friction present, vmax is given by: [ATTACH type="full" alt="formula for vmax.png"]317645[/ATTACH] if we put theta = 90 degrees in the formula above, which is the angle in the well of death, then [ATTACH type="full" alt="formula for vmax when theta is 90.png"]317646[/ATTACH] But that is not acceptable, as vmax can't be an imaginary number. I understand that if theta = 90 degrees, then the minimum value of v i.e vmin is given by: [ATTACH type="full" alt="formula for vmin when theta is 90.png"]317647[/ATTACH] This is understandable. How do I calculate the value of vmax when theta = 90 degrees ? [TABLE] [TR] [TD][ATTACH type="full" alt="5wu0m2DcOd2JT4QJbJ09VhfpppX354mrciT9qm4QX=s40-p-mo.png"]317649[/ATTACH][/TD] [TD]ReplyForward[/TD] [/TR] [/TABLE] [/QUOTE]
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Well of Death: Physics behind it
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