Analysis on the circular motion

1. Dec 14, 2004

UrbanXrisis

I dont know what they are asking. Any tips?

Last edited by a moderator: May 1, 2017
2. Dec 14, 2004

Staff: Mentor

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.

3. Dec 14, 2004

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.

4. Dec 14, 2004

Pyrrhus

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

So on your analysis on the circular motion

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

Last edited: Dec 14, 2004
5. Dec 14, 2004

UrbanXrisis

but how does that get me the height?

6. Dec 14, 2004

Tide

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

HINT: Energy is conserved!

7. Dec 14, 2004

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.