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

[leoflindall]

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 = [tex]\sqrt{2 g r}[/tex] , 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 [tex]\frac{V^{2}}{r}[/tex]

This equated to mg yields,

mg=m[tex]\frac{V^{2}}{r}[/tex]

which rearranges;

V =[tex]\sqrt{gr}[/tex]

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

 

[tiny-tim]

Hi Leo!

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

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? )

 

[leoflindall]

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 [tex]\sqrt{2gr}[/tex], 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

 

[tiny-tim]

Hi Leo!

Yes, the second approach is the way to solve this problem.

… The reason I am confused is that I know that the escape velocity from a planet is [tex]\sqrt{2gr}[/tex], 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!

 

[leoflindall]

Cheers!

Thank you for your help!

Leo

 

[Rayquesto]

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.