Analysis on the circular motion

[UrbanXrisis]

http://home.earthlink.net/~urban-xrisis/phyyy001.jpg

Is the answer just H-R?

I don't know what they are asking. Any tips?

 

[Doc Al]

Do you really think the child loses contact with the slide before even reaching the rounded section? Don't just guess.

Consider that in maintaining contact over the rounded section, the child must undergo circular motion. And that requires a centripetal force. (What force holds the child to the slide?) At some point, the child will be going too fast for the force to maintain the circular motion---off he goes.

 

[Tide]

At some point on the circular section the child's speed will be such that mv^2/r exceeds the normal force holding him on the slide.

 

[Pyrrhus]

When the normal force = 0 the boy loses contact with the surface.

So on your analysis on the circular motion

n - mg \cos \theta = -m \frac{v^2}{R}

 

[UrbanXrisis]

but how does that get me the height?

 

[Tide]

You should be able to determine the speed of the child in terms of her height.

HINT: Energy is conserved!

 

[Pyrrhus]

What i said above means when v^2 = Rg \cos \theta it will be at the point it leaves the surface.

Hint: Use this fact and Tide's hints.