at the top, mv^2/r=mg
r=v^2/g
The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4.6 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times
2 answers
oops. Stinking thinking, I didn't read it.
at the top mv^2/r=mg
but v a the top is the initial v, minus that lost to PEnergy
vf^2=vi^2-2gh
or vf= sqrt (4.6^2-2*9.8*h)
now, to the solution
vf^2/r= g
(4.6^2-2*9.8*2r)^2 =rg
solve for r
at the top mv^2/r=mg
but v a the top is the initial v, minus that lost to PEnergy
vf^2=vi^2-2gh
or vf= sqrt (4.6^2-2*9.8*h)
now, to the solution
vf^2/r= g
(4.6^2-2*9.8*2r)^2 =rg
solve for r