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A roller-coaster car enters the circular-loop part of the ride. At the very top of the circle (where the people in the car are upside down), the speed of the car is v and the acceleration points straight down. If the radius of the loop is r and the total mass of the car (plus passenger) is m, find the magnitude of the normal force exerted by the track on the car at this point. |
Answer» Solution :There are TWO forces acting on the car at its HIGHEST point: the normal force exerted by the track and the gravitational force, both of which point DOWNWARD.  The combination of these two forces, `F_(N)+F_(w)`, provides the centripetal force, so `F_(N)+F_(w)=(mv^(2))/(r)IMPLIES F_(N)=(mv^(2))/(r)-F_(w)`. |
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