Mechanics problem - how fast for a car to take off on a hill.

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Homework Help Overview

The discussion revolves around a mechanics problem involving a stuntman needing to determine the speed required for a car to become airborne at the top of a hill, modeled as a vertical circle with a radius of 100m. The original poster presents two different methods to solve the problem, leading to differing answers.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to solve the problem using two methods: one based on equating kinetic and potential energy, and the other using centripetal force. Some participants question the validity of the first method, suggesting it may not be appropriate for the scenario.

Discussion Status

Some participants provide guidance, indicating that the second approach is more suitable for the problem. There is an exploration of the relationship between the equations derived and the concept of escape velocity, leading to further questions about the assumptions made in the first method.

Contextual Notes

Participants express confusion regarding the distinction between angular velocity and tangential velocity, indicating a need for clarification on these concepts as they relate to the problem.

leoflindall
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Homework Statement



I have solved the following question in two ways which give different answers. Both methods are similar but a little bit different so I think one is wrong. Can anyone advise me which (if either) of the answers is right?

A stuntman is preparing his next stunt. He needs to get his car airborne once he reaches the top of the hill. The hilltop is roughly a vertical circle with radius 100m. How fast does he need to drive.


Homework Equations


E=1/2 m v^2
PE=mgh
Centripetal acceleration (a) = (V^2)/r


The Attempt at a Solution



Solution 1.

At the point of takeoff the energy of the car must be equal to the potential energy at the top oh the hill. So we can say that KE=PE, and thus

1/2 mv^2 = mgh

which rearranged:

V = \sqrt{2 g r} , where r is the radius of the hill, and g is the acceleration.

This gives an answer of 99.6 mpm.

Solution 2

At the point of take off the Force acting upwards on the car must be equal to the weight (mg) of the car. Knowing the formula for centripetal acceleration we can derive an equation for the force upwards,

F = m \frac{V^{2}}{r}

This equated to mg yields,

mg=m\frac{V^{2}}{r}

which rearranges;

V =\sqrt{gr}

which gives an answer of 70.4mph.




Both answers are plausible as there is only a factor of root 2 difference. I would be very appreciative if anyone could advise me as to which method is correct, and what is wrong with the other!

Leo
 
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Hi Leo! :smile:

(have a square-root: √ and try using the X2 icon just above the Reply box :wink:)

Your 1 has nothing to do with the problem … it's the equation for the speed at the bottom if it starts from rest at the top of the circle!

(Your 2, correctly, has no height difference or KE difference … why does your 1 have both? :confused:)
 
Hi tiny-tim,

I thought the second one was right. The reason I am confused is that I know that the escape velocity from a planet is \sqrt{2gr}, and having noticed the similarity between the equation (2) I derived I wondered if I had made a mistake.

So, the second approach is the best way to solve this problem?

Many Thanks

Leo
 
Hi Leo! :smile:

Yes, the second approach is the way to solve this problem. :wink:
leoflindall said:
… The reason I am confused is that I know that the escape velocity from a planet is \sqrt{2gr}, and having noticed the similarity between the equation (2) I derived I wondered if I had made a mistake.

Using dimensional analysis (units), any speed that depends on g and r has to be a multiple of √(gr), so this similarity is to be expected! :smile:
 
Cheers!

Thank you for your help!

Leo
 
this problem is confusing to me. I am really sorry that I cannot give much insight to the problem, but it sounds to me like what they are asking is what angular velocity must the car travel to overcome the centripetal acceleration. God this problem is confusing. But I've also notice that angular velocity and tangential velocity are different. So, god sorry. Please message me to tell me how you figured this out. I really want to know.
 

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