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.

Well, let's take a look at this. The child starts at point P and slides down the straight section of the slide. Gravity pulls the child down, but there is a normal force from the slide pushing back up to counteract gravity. As the child reaches the circular arc, the normal force decreases because the surface of the slide curves away. Eventually, the normal force reaches zero and the child becomes airborne.

To find the height at which this happens, we need to consider the forces acting on the child at that point. In this situation, the only vertical force acting on the child is gravity. So, at the point of becoming airborne, the force due to gravity must be equal to the weight of the child.

Let's call the height at which the child becomes airborne h. At this point, the child is no longer in contact with the slide, so the normal force is zero. This means the only vertical force acting on the child is gravity, which is equal to the weight of the child.

Therefore, we can set up the equation:

mg = 0

Solving for h, we find that the height at which the child becomes airborne is h = 0.

Well, that's a bit anticlimactic, isn't it? It seems like the child becomes airborne at ground level. I guess this poorly designed playground slide has a very short lifespan!

To find the height at which the child becomes airborne, we can use the principle of conservation of energy. At the point where the child loses contact with the ramp, all of their energy will be converted to potential energy.

First, let's consider the energy at the starting position, point P. At this point, the child is at a height h above the ground. The potential energy at this point is given by mgh, where m is the child's mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Next, let's consider the energy at the point where the child becomes airborne on the circular arc. At this point, the child's potential energy is zero since they are in mid-air. However, we need to calculate the child's kinetic energy at this point. The kinetic energy is given by (1/2)mv^2, where v is the child's velocity.

Since the slide is frictionless, we can equate the initial potential energy to the final kinetic energy:

mgh = (1/2)mv^2

Simplifying and canceling out the mass:

gh = (1/2)v^2

We can express the velocity v in terms of the height h using the relationship between velocity and centripetal acceleration for circular motion. The centripetal acceleration is given by (v^2)/r, where r is the radius of the circular arc. At the point where the child loses contact with the ramp, the centripetal acceleration is equal to the gravitational acceleration g, since the normal force goes to zero. Therefore:

(v^2)/r = g

Solving for v and substituting it back into the previous equation:

gh = (1/2)((rg)^2)/r
gh = (1/2)(r^2)g
h = (1/2)r

Therefore, the height at which the child becomes airborne is half the radius of the circular arc.

Note: In order to give a specific answer in units of meters, the radius of the circular arc needs to be provided.

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?

relate 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.

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(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.