Solution to AP Chapt 7 prob 55

Solution to Chapt 7 Problem 55 b

Remember that sin theta = r/L and cos theta is h/L [r is radius or Dx and h is height or Dy) and that theta is angle with y axis here.

Then we have for y direction: Ft cos theta =Ft h/L =Mg and for x direction: Ft sin theta = Ft r/L=Fc = M v^2/r We are given v and L ( and M but it cancels out here) so h r and theta are unknowns. Given h or r we can find theta from inverse cos or sin.

We have from y direction Ft = Mg/(h/L) and from x direction Ft = v^2/(r^2/L) setting the two equations for Ft equal to each other we have: Mg/(h/L) =M v^2/(r^2/L) and the Ls and Ms cancel so g/h = v^2/r^2

By Pythagorean Theorem r^2=L^2-h^2 so now we have g/h = v^2/(L^2-h^2).
Since we are after h, lets invert the equation and get h/g= (L^2-h^2) / v^2
multiply both sides by v^2 and collecting terms on one side as for a quadratic equation: h^2+(v^2/g)h –L^2= 0
the solution to this quadratic is [- v^2/g+/- squareroot(v^4/g^2 + 4 L^2)}/2

Putting numbers in this awful mess ( and I’ll use 9.8 here even though it hurts me to do so) we get h [-16 /9.8 +/- sqrt(256/96+9)]/2= [-1.63 + sqrt (11.67)]/2 = 1.79/2 = 0.893; h/L =cos theta = .595 and theta = 53.5 degrees.
and r = 1.5 sin 53.5 = 1.21m

Lets check

does Fc /(sin 53.5)= Ft = Mg/cos 53.5 ?

v^2 is 16 (M v^2/r )/sin 53.5= (16/1.21)/.804= 16.5 M
Mg/cos 53.5 = 1.68 Mg = 16.5 M so we are right but there must be a better way to do this problem