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April 24, 2014

April 24, 2014

Posted by **Genevieve** on Friday, October 8, 2010 at 2:14pm.

img716.imageshack.us/img716/8381/25746193.jpg

A child starts at point P and slides down both sections of the slide. At some point on the circular arc, the normal force goes to zero and the child loses contact with the ramp.

Assuming the forces of friction are negligible, at what height from the ground will the child become airborne?

Answer in units of m.

- Physics-Mechanics -
**bobpursley**, Friday, October 8, 2010 at 3:51pmrelate the change in PE from the vertical fall to changes in KE

KE change: 1/2 m v^2

height change: 4.9-radius+ radius(1-CosTheta)

PE change: mg*height change

check that

where theta is the angle measured from the point on the circle to the center and the x axis.

Now, the object flies off when the normal component of weight (mgSinTheta) is equal to centripetal force (mv^2/r)

set the equal.

from setting PE change=KE change, you should be able to solve for v^2. Put that into the equation with the forces, and solve for angle.

- Physics-Mechanics -
**Genevieve**, Friday, October 8, 2010 at 4:02pmHi,

I am confused on the height change.

By 4.9 did you mean 4.8? and if so do I set it equal to zero to solve?

is the v in the centripetal force different from the v in KE?

- Physics-Mechanics -
**bobpursley**, Friday, October 8, 2010 at 4:55pmyes, 4.8 I was using my memory of the pic as 4.9

KEchange=PE change

1/2 m v^2=mg(4.**8**- radius+ radius(1-cosTheta)

solve for v^2

then put that v^2 in this:

mv^2/r=mgSinTheta

and solve for Theta

You may have some tricky algebra.

It is the same v in KE, and force.

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