# Circular Motion, Did I do it right?

by bob1182006
Tags: circular, motion
 P: 492 Halliday & Resnick 5e Chapter 5 Problem 14 1. The problem statement, all variables and given/known data A car moves at some constant speed on a straight but hilly road. One section has a crest and dip of the same 250-m radius. (a). as the car passes over the crest, the normal force on the car is 1/2 the 16-kN weight of the car. What will be the normal force on the car as it passes through the bottom of the dip? (b). What is the greatest seed at which the car can move without leaving the road at the top of the hill? (c). Moving at the speed found in (b) what will be the normal force on the car as it moves through the bottom of the dip? 2. Relevant equations Fnet=ma a=v^2/R 3. The attempt at a solution a. on the crest: Normal Force up, weight down, acceleration down N-mg=-ma $$\frac{-N}{m}+g=\frac{v^2}{R}$$ N=.5mg $$\frac{N}{m}=\frac{.5mg}{m}$$ N/m=.5g $$R(-.5g+g)=v^2$$ $$v=\sqrt{.5gR}$$ v=35 m/s v constant so can use it to find N during the dip. on the dip: Normal Force up, weight down, Acceleration Up N-mg=ma N=m(g+a) $$N=m(g+\frac{v^2}{R}$$ N=1/2mg m=2N/g $$N=\frac{2N}{g}(g+\frac{v^2}{R}$$ plugging in I get N=24N b. The greatest v that the car can reach will be enough to give N=0? $$x: ma_x=0$$ $$y: N-mg=-ma_y$$ $$N-mg=-m\frac{v^2}{R}$$ $$g=\frac{v^2}{R}$$ $$v^2=gR$$ $$|v|=\sqrt{gR}$$ plugging in v=50 m/s c. $$x: ma_x=0$$ $$y: N-mg=ma_y$$ $$N=m(a_y+g)$$ $$N=m(\frac{v^2}{R}+g)$$ $$N=\frac{W}{g}(\frac{v^2}{R}+g)$$ $$N=\frac{Wv^2}{gR}+W$$ plugging in I get N=32.3 kN Thanks in advance, was reviewing and couldn't repeat my reasoning on this problem :s.