Note the x-y coordinate system of the applet. The origin is at the center of the MGR, +x is to the right, and +y is up. We'll use the following symbols for brevity:

R = radius of merry-go-round (fixed at 10 m)

T = period of merry-go-round

V_{o} = initial throw speed of ball relative to Blue rider

q = initial throw direction of ball (0° is toward center of
MGR, +90° is in +y direction)

V_{r} = speed of either rider relative to the ground* (Note that V_{r =} 2pR / T)

*For simplicity, we'll sometimes refer to the Inertial Frame as the "ground".

*In order to help keep Frames of Reference straight, we'll
change to blue italics when discussing the Non-Inertial Frame, that is, the frame that
includes the MGR and its riders.*

1. Click on Initialize/Reset to set the parameters to their original
values. Toggle to the Inertial
Frame, if necessary. Now click Forward. Note that even though Blue threw the
ball toward the center of the MGR, the direction of the ball's path is diagonally toward
the top of the screen. That's because the ball has, in addition to an x-component of
velocity, a y-component equal to Blue's velocity at the instant of the throw. These
components are designated V_{x} and V_{y} in the applet window. Note
that V_{y} = V_{r} = 2p(10 m) / (5 s) = 12.56
m/s. Notice also that the path of the ball is a straight line, since V_{x}
and V_{y} do not change. That makes sense because there's no force acting on
the ball after it leaves Blue's hand.

2. Now Pause and Reset the applet, keeping the same values of T, V_{o},
and q. *Toggle to the
Non-Inertial Frame. Try to predict what path the ball will take from this reference
frame. Remember, this is like the path seen by a rider on the MGR. When you're
ready, click Forward. Why doesn't the MGR move? (The riders view the rest of
the world as moving while they are stationary.) Notice that the ball starts out
heading directly for the center of the MGR but then veers away. It seems as if
there is a force accelerating the ball away from the MGR.*
Yet, we saw for the Inertial Frame that there was no force on the ball. From
an Inertial point of view, we can explain the *apparent* force on the ball seen by
the riders as being due to their own rotation.

3. Pause the applet, increase V_{o} to 50 m/s and Reset.
Toggle to the Inertial Frame. What do you expect the path of the ball to be this
time? (The V_{y} will be the same as before since Blue's speed has not
changed. However, the V_{x} is 5 times greater than before.) Start the
applet. Were you correct? The ball barely misses the center of the MGR.
(Do you see why the ball cannot pass directly through the center as long as q = 0°?) *Now what would this look
like from the Non-Inertial Frame? Let's predict first. We know the ball will
miss the center of the MGR. That result has to be the same in both frames. In
the Non-Inertial Frame, however, the ball's initial velocity vector is directly toward the
center. Therefore, if the ball is to miss, it will have to veer away. Try it
and see.*

4. What happens when you swing a ball in a circle at the end of a tether and
suddenly let go? The ball travels tangent to the circular path that the ball had.
The applet can demonstrate this. Set V_{o} to 0. This means
that Blue simply lets go of the ball. Reset the applet, toggle to the Inertial
Frame, and click Forward. *As the motion progresses, try to
imagine what the path of the ball would look like from the Non-Inertial Frame. Then
Pause, Reset, toggle to the Non-Inertial Frame, and see if you're right.*

5. Pause the applet and toggle to the Inertial Frame. What would V_{o}
and q have to be so that the ball wouldn't move from its
initial position at (x,y) = (R,0)? That is, we want the velocity of the ball
relative to the ground to be 0. Enter your predicted values, Reset, and click
Forward. Were you right? Hopefully, you realized that the velocity of the ball
relative to Blue would have to be the opposite of Blue's initial velocity relative to the
ground. The latter is 12.56 m/s in the +y direction. Therefore, the ball's
velocity relative to Blue must be 12.56 m/s in the -y direction (q
= 90°). *Now predict what the path would look like from
the Non-Inertial Frame and try it.*

*If you're ready, go on to some quantitative problems
...*