# Help me solve this using alternative approaches -- car driving around a circular track

Tags:
1. Jul 30, 2017

1. The problem statement, all variables and given/known data
A car of mass m is initially stationary on a smooth track at distance h above the ground. What would be the minimum value of h required in order for the car to remain on the track throughout its journey around the loop of radius r?

2. Relevant equations
KE(bottom)>PE (top)
F=mv^/r
3. The attempt at a solution
I solved this using the minimum velocty, sqrtgr, at the top and applying energy conservation. But I tried this using two other alternate approaches that didmn't yield the correct results:
1)KEbottom>PE(Top)= mv^2(.5)=mg(2r). This gives the minimum velocity as sqrt4gr. Although it should be sqrtgr according to centripetal force eq (mv^2/r=mg).
2) Second approach: I have attached the snap below of the solution.Now for h to be minimum, costheeta should be one. Why is it taken to be -1.

Kindly resolve both of these queries.

#### Attached Files:

• ###### Screenshot_2.png
File size:
48.4 KB
Views:
18
Last edited by a moderator: Jul 31, 2017
2. Jul 30, 2017

### haruspex

Velcity where? You seem to have calculated the velocity at the bottom which corresponds to the PE at the top. But it is not stationary at the top.
Yes, but what you want is the minimum h which satisfies the condition for all theta, so you want the theta which maximises the demand on h.

Last edited: Jul 31, 2017
3. Jul 31, 2017

1) I got you there. But why aren't we able to solve using the inequality that if the car completes the vertical circle, its speed is greater than sqrt4gr. The answer still doesn't turn out to be correct. Why is this the case?
2) I didn't get your point. Could you please simplify it or explain some other way?

4. Jul 31, 2017

### haruspex

My difficulty here is that I do not understand your reasoning in post #1. Can you spell out it out in more detail please, being quite specific about the points in the process that you are calculating the velocities and energies for?
It's late here... I'll try tomorrow.

5. Jul 31, 2017

### haruspex

You have h≥r(1-(3/2)cos(θ)) as the condition for staying on the track as far θ. To stay on the track for the whole loop the condition has to be true for all θ. The range of cos(θ) is -1 to 1, so you need h≥r(1-(3/2)cos(θ)) ∀ cos(θ). The value of cos(θ) which maximises the right hand side is -1, so for the condition to be true ∀ cos(θ) that is the critical case that h needs to satisfy.

6. Aug 1, 2017

That is my question. We need the minimum h, why are we maximizing the RHS?

7. Aug 1, 2017

### haruspex

Yes, but it is the minimum h which satisfies a certain condition. That condition is that it exceeds the RHS for all possible values of θ. That is quite different from finding the minimum value of h which exceeds the RHS for some value of θ.
Think about an analogy. An aircraft is to fly a level course over a mountain range. The height of the terrain can be expressed as y=y(x), where x is the horizontal distance. If the aircraft is to fly at height h then you want h≥y(x) ∀x. It is not enough that h≥y(x) for some x.