The Inertial Frame may be referred to as the "ground". The Non-Inertial Frame includes the MGR and the two riders.
The problem difficulty is indicated by the number of stars.
*1. Click on Initialize/Rest to set the parameters to their original values. Toggle to the Inertial Frame if necessary.
*2. Your problem now is to throw the ball so that it passes through the center of the MGR. This will require a negative q, of course, but you must determine exactly what the angle is.
**3. In problem 2, you probably used the fact that the y-component of the ball's velocity relative to Blue had to be the opposite of the y-component of Blue's velocity relative to the ground at the instant the ball was thrown. In this way, the y-component of the ball's velocity relative to the ground would be 0. That is, at t = 0,
Vy,ball wrt ground = Vy,ball wrt Blue + Vy,Blue wrt ground ("wrt" = "with respect to")
0 = Vy,ball wrt Blue + Vy,Blue wrt ground
Vy,ball wrt Blue = -Vy,Blue wrt ground
**4. With T = 2.5 seconds, what must Vo and q be so that Red catches the ball after one full rotation? Test your prediction. Look at it from the Non-Inertial Frame, too. How do Vo and q depend on T?
**5. With T = 5 seconds, what must Vo and q be so that Red catches the ball after one quarter rotation? (The ball cannot pass through the center in this case.)
***6. Suppose Red is to catch the ball after rotating through any angle a less than p/2.
- Determine formulas for Vo(a) and q(a). Then check them for particular values of a.
- In a real situation on the MGR where one rider wants to throw the ball to the other, a will generally be small. Using the small angle approximations, sina = a and cosa = 1, eliminate a from the equations from part a in order to obtain Vo as a function of q only. Then use the applet to investigate the range of angles, q, for which the approximation works well.