Conservation of energy I think?

Daria Verfaillie: Use both conservation of energy AND circular motion analysis.Believe it or not, it will not matter what Earth's gravity is, what the body's mass is, or what the dome radius is. We will use g, m, and R to represent these, but they will cancel out of the equation eventually.This problem requires a lot of faith in Algebra.Define GPE = 0 at base of dome.Energy at top:GPEtop = m*g*RKEtop = 0GPE at angle theta from the top:GPEangle = m*g*R*cos(theta)KEangle = 1/2*m*v^2Kinetic energy at angle theta:KEangle = GPEtop - GPEangleThus:1/2*m*v^2 = m*g*R*(1 - cos(theta))Cancel dependence on mass and solve for v^2:1/2*v^2 = g*R*(1 - cos(theta))v^2 = 2*g*R*(1 - cos(theta))What does it mean for the body to be "in contact" with the dome? It means that the normal force from the dome on the body has a value greater than zero. At the instant of loss of contact, N = 0.Forces acting on the body at angle theta:Locally Perpendicular to dome:N: outwardm*g*cos(theta): inwardLoc! ally parallel to the dome:m*g*sin(theta), downwardWe don't really care about parallel to the dome, because all that controls is the tangential acceleration.Newton's 2nd law for perpendicular to dome:m*g*cos(theta) - N = m*a_centripetalSubstitute expression for centripetal acceleration:m*g*cos(theta) - N = m*v^2/RSubstitute v^2:m*g*cos(theta) - N = m*(2*g*R*(1 - cos(theta)))/RCancel dependence on R:m*g*cos(theta) - N = m*(2*g*(1 - cos(theta)))Set N = 0 at the instant of contact loss:m*g*cos(theta) = m*(2*g*(1 - cos(theta)))Cancel dependence on m and g:cos(theta) = (2*(1 - cos(theta)))Call cos(theta) C, and substitute:C = (2*(1 - C))Solve for C:C = 2 - 2*C3*C = 2C = 2/3Thus:cos(theta) = 2/3Take inverse cosine:theta = arccos(2/3)which evaluates to theta = 48.2 degrees...Show more