 #1
 815
 95
Homework Statement:

A horizontal disk rotates at an angular speed ##\omega_1## clockwise around its center of mass (which is where the origin of the noninertial frame lies). A person with a rope is turning counterclockwise at an angular speed ##\omega_2## on that disk. The rope has length ##l_2##. The man himself is at a distance ##l_1## from the origin of the disk and the rope also moves at a height ##h## above the horizontal disk. A stone is attached to the end of that rope, which becomes a pendulum. At ##t=0##, the pendulum is maximally far away from the origin (thus, the origin, the man and the stone stand on a straight line).
a) Find an expression for the speed ##v(t)## of the stone at a given moment ##t## in terms of the Cartesian coordinate system at rest. Choose the origin of such a system to be located at the center of the disk. Choose the ##z## axis to be vertically upwards, the ##x## axis according to the straight line that goes through the stone at ##t=0## and the ##y## axis perpendicular to the previous ones.
b) Suppose the man releases the stone at ##t=0##. Where does the stone fall on the disk?
NOTE:This exercise has been translated from Dutch by Google Translator. Let me know if there's something unclear and we can all discuss about it.
Relevant Equations:
 Please see first picture
Where:
1) ##A## is the translational acceleration, ##\Omega## the angular velocity and ##\dot \Omega## the angular acceleration (all relative to the inertial frame attached to the ground ##F##).
2) ##r'##, ##v'## and ##a'## are the position, velocity and acceleration vectors, all relative to the frame attached to the roundabout ##F'## (and thus ' has nothing to do with derivatives on above equation).
 Picture of the problem (imagine the man has a pendulum):

a) I think the exercise is just asking us to find ##v(t)## of the stone with respect to the inertial frame ##F##.
If the pendulum is not released, then its velocity will be same that the velocity of the man. Thus:
Expression for ##v(t)## with respect to the inertial frame F:
$$\omega_s=\omega_m−\omega_d$$
$$\frac{v_s}{l_1 + l_2} = \frac{v_m}{l_1}  \frac{v_d}{l_2}$$
Thus the velocity of the stone is:
$$v_s = (l_1 + l_2) \frac{Rv_m + l_1 v_d}{R l_1}$$
But this velocity is not in function of ##t##, so I guess this is not what the exercise is asking for.
Besides I am not convinced that the assumption I made above is correct (If the pendulum is not released, then its velocity will be same that the velocity of the man)...
b) Here I guess the exercise is asking for a qualitative answer. I expect the pendulum when released to have the direction of the black arrow:
Attachments

88 KB Views: 49
Last edited: