Calculus with Physics question

[CanadaAotS]

Hello all! first of all I would like to say this is NOT homework , I know that's probably what everyone says but its true. I actually thought up this question myself (yes I have time on my hands ) but have not been able to solve it. I'm in Calculus right now, but finished taking grade 12 physics last year. Needless to say I'm a bit rusty, which is why I need help on answering this...

 

I'm not sure if anyone here knows what I'm talking about but the playground near my house has a big concrete cylinder thing to play in. This problem makes more sense if you keep in mind lol

 

A 35 kg child in a 3 m radius cylinder concrete thing ( ) slides down from the left side at the center point. The coefficient of friction between the child and 'pipe' is 1/8. How fast is he travelling when he hits the bottom of the 'pipe'.

 

diagram should be attached to this post.

 

I figured out that the bottom half of the pipe is all you need for the question and the equation of the bottom of a half circle is [math]y = -\sqrt{r^2 - x^2}[/math]. r^2 is 9 in this question.

 

I know I need calculus for this, because the net force on the kid is constantly changing. As he slides down the slope (derivative) is changing causing the friction to change as well.

 

What I can't get my head around is how I use this information to solve the problem, it's probably just that I'm forgetting some physics equations...

 

btw the derivative of [math]y = -\sqrt{3^2 - x^2}[/math] is [math]y' = \frac{-2x}{\sqrt{3^2 - x^2}}[/math]

 

thanks in advance!

[5614]

Don't you need the height he has fallen through? As in he starts at 1m and ends up at ground level.

 

If the slope has a constant gradient then he is accelerating at [math]a = mg sin \theta[/math] where:

m = boy's mass

g = gravity ie. 9.8

[math]\theta[/math] = angle between the slope and the horizontal.

[□h=-16πT]

Your derivative is wrong by the way.

 

[math]y=-\sqrt{9-x^2} [/math]

[math]y'=-\frac{1}{2}(9-x^2)^{-\frac{1}{2}}\frac{d(-x^2)}{dx}=\frac{x}{\sqrt{9-x^2}}[/math]

 

I can't remember how you do coefficient of friction and all the gubbins, but the problem without coefficients of friction just requires that the kid's kinetic energy at the bottom of the pipe be the same as its potential energy at its initial position, height=3. Can you not integrate friction coefficients into conservation of energy.

[CanadaAotS]

wha? where does the 1/2 in the front come from?

[□h=-16πT]

You've got

 

[math](9-x^2)^{\tfrac{1}{2}}[/math]

 

Then you use the chain rule.

[[Tycho?]]

wha? where does the 1/2 in the front come from?

 

A square root is the same as an exponent to one half.

 

Square root of 4 = 4^(1/2)

 

Its just a different way of writing it, makes derivatives easier.

[CPL.Luke]

well the magnitude of fn is going to be mg cos theta

 

where theta is the angle formed by the tangent line of the cylinder and the gravity vector, if you come up wiht an equation for theta the rest of the problem should be easy, because if you integrate the force do to friction with respect to the distance it acts through, then you have the work done by friction, add that to the work done by gravity, then plug into the equation for kinetic energy and solve for v and your good.

 

it may be easier to do the problem out in polar co-ordinates but i'm not sure

 

when you mean grade 12 physics do you mean ap physics c, ap physics b, or physics one?