Posted by Genevieve on .
A poorly designed playground slide begins with a straight section and ends with a circular arc as shown in the figure below.
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.

PhysicsMechanics 
bobpursley,
relate the change in PE from the vertical fall to changes in KE
KE change: 1/2 m v^2
height change: 4.9radius+ radius(1CosTheta)
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. 
PhysicsMechanics 
Genevieve,
Hi,
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? 
PhysicsMechanics 
bobpursley,
yes, 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(1cosTheta)
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.