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- Homework Statement
- 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 ?

- Relevant Equations
- $$v_{max} = \sqrt{\frac{rg(sin\theta +\mu cos\theta)}{(cos\theta -\mu sin\theta)}}$$

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:

if we put theta = 90 degrees in the formula above, which is the angle in the well of death, then

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:

This is understandable.

How do I calculate the value of vmax when theta = 90 degrees ?

On a banked road with friction present, vmax is given by:

if we put theta = 90 degrees in the formula above, which is the angle in the well of death, then

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:

This is understandable.

How do I calculate the value of vmax when theta = 90 degrees ?

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