Maximum uniform speed on a arc of a circular path ?

[Buffu]

Question :-
A car has to move on a path, that is a arc of a circle of radius ($R$). The length of the path is ($L$). Suppose it starts on the highest point of the path, find the highest uniform speed for which, it does not lose contact with the path on any point ?

My attempt :-
I made a free body diagram, http://imgur.com/uwOokiJ
(The direct image insertion is not working, since this is my first post i don't know how to fix it . So i will leave a link to image, please check it http://imgur.com/uwOokiJ )

From the diagram i got,
The maximum permitted velocity at each point would be
$\large{mv \over R^2} = \large{mg \cos(\theta)}$
from which i got,
$v = \sqrt{Rg \cos(\theta)}$

Since we need to find maximum value of $v$, thus we need to find maximum of $\cos(\theta)$, which is $1$.

So the answer is $v = \sqrt{Rg}$

Which is incorrect.
correct answer is $\sqrt{Rg\cos\left({l\over 2R}\right)}$.

I think i am close but, could not get it. Please help me.

 

[kuruman]

Hi Buffu and welcome to PF.

Your answer is the speed at which the car loses contact immediately at the top of the arc. You want the car to stay in contact all the way down to the end of the arc. What does your expression $v = \sqrt{Rg \cos(\theta)}$ mean? How did you get it?

 

[haruspex]

This is a minimax problem. You found the max speed at a given θ. You want the max speed that does not exceed that for any θ in a certain range.

 

[Buffu]

Hi Buffu and welcome to PF.

Your answer is the speed at which the car loses contact immediately at the top of the arc. You want the car to stay in contact all the way down to the end of the arc. What does your expression $v = \sqrt{Rg \cos(\theta)}$ mean? How did you get it?

We know, from the diagram that ${mv^2 \over R } = mg\cos(\theta) - N \qquad (1)$

Now maximum speed just before the car loses contact will be same as the minimum speed at which car loses contact.
If car loses contact then $N = 0$
Thus $(1)$ becomes ${mv^2 \over R } = mg\cos(\theta) \qquad (3)$

${mv^2 \over R} = mg\cos(\theta)$
$v = \sqrt{Rg\cos(\theta)}$

So i think it means that the maximum permitted speed at any point is $\sqrt{Rg\cos(\theta)}$ for $\theta$ being the acute angle between $mg$ and $N$(if we extend it towards the center).

 

[Buffu]

This is a minimax problem. You found the max speed at a given θ. You want the max speed that does not exceed that for any θ in a certain range.

I found that to be $\sqrt{Rg}$.
I have be thinking about this for very long but can't find the answer. :(

 

[haruspex]

I found that to be √Rg.

No, that was using the max value of cos(θ). You need v to be below √(Rgcos(θ)) for all θ in some range, so you need to look for the least cos(θ) in that range.

 

[kuruman]

So i think it means that the maximum permitted speed at any point is $\sqrt{Rg\cos(\theta)}$ for θ being the acute angle between mg and N (if we extend it towards the center).

Yes, that is the maximum speed beyond which the car will fly off at some point on the arc. You want the car not to fly off anywhere along the arc. You don't care if the car flies off at the end of the arc. So what do you think angle θ should be in your expression?

 

[Buffu]

No, that was using the max value of cos(θ). You need v to be below √(Rgcos(θ)) for all θ in some range, so you need to look for the least cos(θ) in that range.

No, that was using the max value of cos(θ). You need v to be below √(Rgcos(θ)) for all θ in some range, so you need to look for the least cos(θ) in that range.

Oh that means i need to find the highest value of $\theta$, since $\cos(\theta)$ is decreasing in first quadrant. Thanks.

 

[Buffu]

Yes, that is the maximum speed beyond which the car will fly off at some point on the arc. You want the car not to fly off anywhere along the arc. You don't care if the car flies off at the end of the arc. So what do you think angle θ should be in your expression?

Got it thanks for the help.