(II) Two blocks, with masses $m_A$ and $m_B$ are connected to each other and to a central post by thin rods as shown in Fig. 5-41. The blocks revolve about the post at the same frequency $f$ (revolutions per second) on a frictionless horizontal surface at distances $r_A$ and $r_B$ from the post. Derive an algebraic expression for the tension in each rod.
(II) A 975-kg sports car (including driver) crosses the rounded top of a hill (radius $=$ 88.0 m) at 18.0 m/s. Determine ($a$) the normal force exerted by the road on the car, ($b$) the normal force exerted by the car on the 62.0-kg driver, and ($c$) the car speed at which the normal force on the driver equals zero.
the situation in this question is the following. This car is moving to the right until it meets a hell which is inclined at 18 degrees with respect the horizontal. This car is moving at 95 kilometers per hour. We then have to discover what is the minimal radios off this transition section such that the car do not leave the road. So these is the radios, it's the radios off the circle and we have to discover what is the minimum value off are such that the car do not fly away went passing through this transition. For that, let us analyze closer. Why is this transition region so lives as room it in the transition region? Notice that the transition is nothing else than a section off a circle. So during the transition the car is under giving a circular motion. Then we can treat it as usual. The condition that the car never leaves the road translates to the condition that the normal is bigger than or at most equals 20 So the condition is the following. The normal forced must be bigger than or at most equals 20 But now what does the normal force has to do with the radios off the circle. For that, you have to remember a few beats about this and typical force. So whenever an object is undergoing a surgical emotion, but there is a centripetal force acting on that object on that centripetal force is given by the mask off the object times the velocity off that object squared, divided by the radios off the trajectory. But then what does this centripetal force has to do? The normal force. It happens that the centripetal force is the net force that is acting on the radio, their action, which is the direction that passed through the center off the trajectory. Then you can see that in any situation, the normal force is part off the centripetal force. But obviously the normal forest isn't the wolfing. There is actually a component off the weight force that is composing the centripetal force. Let me call this component the radio component off the weight force. And it is like this does this level. You are the other component off the weight force Would be these one, then genteel component. Wait, I'm calling the beauty. I noticed that this Ando is the same Anno as these one because we have to parlor lines that are crossed, my, uh, straight line. Then we can also say that this angle is equal to teeter and then we are able to relate the weight with the radio component off the weight. It was in that triangle we have the following. This is the triangle. And then these is the 90 degree angle is the hypotenuse business WR and this is W t these this Tita, you can see that the co sign off Peter in that context is given by the address in sight. So the radio component divided by their party news which is the full weight then the radio component is equal to the weight times the curve. Sign off, Tita. The reform. We can write this centripetal force as follows at any point off this conference, this centripetal force is given by the radio component off the weight force which is pointing to the center. Sure, it's positive. Mind is the normal force that is pointing outwards. These easy. It goes to the mass times velocity squared, divided by the radios. Then in the limiting situation where the normal is very very close to zero so that we can say that it is effectively zero. We can say the following the radio component off the weight force must be equal to M B squared, divided by r So the radius must be equals two m v squared divided by the radio component off the weight force Using the equation for the radio component off the weight force that we had just directed. We get the following em These weird divided by W times The co sign off detail Remember that the weight force is given by the mass off the object times the acceleration of gravity So these can be written as m v squared, divided by AM G times Nico's sign off, Tita, We can simplify the masses and then we get an equation for the radios. The create parade Ian's is V squared, divided by G times. Niccum, sign off, Tita, We already knew what his V we knew What is G? Actually we remember that he is approximately 9.8 meters per second squared. So the only thing that we are left to discover is the value off this angle, Tita, for that we have to use the geometry of the situation. Notice the following this angle between these lines and these line is 18 degrees. But the angle between the black line on this red line is 90 degrees. I can extend this line so that it's easier to see. As you can see, there is a 90 degree angle here them. With that in mind, we can better mind. What is this angle? Because we know that this world angle is 90 degrees, then these angle must be 90 minus 18 degrees, which results in 72 degrees. These is this angle, but notice that the angle between the radio direction on this blue line is also 90 degrees. So these angle is an angle off 19 Reese. Therefore, we conclude that teeter plus 72 degrees must be equals to 90 degrees on. The only possibility for the result is that Tito is also equals to 18 degrees. Now that we know teeter, we can finish the question So the radios is given by 95 kilometers per hour. We should covert these 95 kilometers per hour two meters per second. For that, we divide it by 3.6. So the velocity is 95 divided by 3.6 meters per second and this is squared and then we divide by 9.8 times. They could sign off 18 degrees and these results in the radios off approximately 75 meters, which is the minimal radios off the transition region.