always going around with the same angular velocity w, so that the angular mo- mentum is L=matang marr=mwr when the mass is close to the center, it has relatively little angular momentum but if we move it to a new position farther out, if we increase r, m has more angular momentum, so a torque must be exerted in order to move it along the radius. o walk along the radius in a carousel, one has to lean over and push sidewise Try it sometime. The torque that is required is the rate of change of L with time as m moves along the radius. If m moves only along the radius, omega stays con- stant, so that the torque is F dl d (mar) dr d t where Fc is the Coriolis force. What we really want to know is what sidewise force has to be exerted by Moe in order to move m out at speed vr= dr/dt. This is Fc=T/r=2mwvr Now that we have a formula for the Coriolis force. let us look at the situation a little more carefully, to see whether we can understand the origin of this force from a more elementary point of view. We note that the Coriolis force is the same at every radius, and is evidently present even at the origin! But it is especiall easy to understand it at the origin, just by looking at what happens from the in- ertial system of Joe, who is standing on the ground. Figure 19-4 shows three successive views of m just as it passes the origin at t =0. Because of the rotati of the carousel, we see that m does not move in a straight line, but in a curved path Fig. 19-4. Three successive views of there must be a force to accelerate it in absolute space. This is the Coriolis force. turntable oving radially on a rotating tangent to a diameter of the carousel where r=0. In order for m to go in a curve, a point m This is not the only case in which the Coriolis force occurs. We can also a circle, there is also a Coriolis force. Why? Moe sees a velocity um around ls circle. On the other hand, Joe sees m going around the circle with the velocity vJ=wM ar, because m is also carried by the carousel. Therefore we know what the force really is, namely, the total centripetal force due to the velocity uJ, or wj/r; that is the actual force. Now from Moe's point of view, this centripetal force has three pieces. We may write it all out as follows Now, Fr is the force that Moe would see. Let us try to understand it. Would Moe appreciate the first term? "Yes, " he would say, even if I were not turning, there would be a centripetal force if I were to run around a circle with velocity uM This is simply the centripetal force that Moe would expect, having nothing to do with rotation. In addition, Moe is quite aware that there is another centripetal force that would act even on objects which are standing stiil on his carousel. This is the third term. But there is another term in addition to these, namely the second term, which is again 2mav. The Coriolis force Fc was tangential when the velocity was radial, and now it is radial when the velocity is tangential. In fact, one ex pression has a minus sign relative to the other. The force is always in the same direction, relative to the velocity, no matter in which direction the velocity The force is at right angles to the velocity, and of magnitude 2mwv