Question

Figure 9-56

A uniform solid sphere of radius r=8.60 cm starts from rest at a height h and rolls without slipping along the loop-the-loop track of radius R=42.00 cm as shown in Figure 9-56. What is the smallest value of h for which the sphere will not leave the track at the top of the loop? (h is measured from the center of the ball at the top of the ramp to the center of the ball at the bottom of the ramp.)

Answer 1

we know,
the smallest speed of any body at the top of the loop must be, v = sqrt(5*g*R)
let w is the angular speed of the sphere when it is at the top of the loop.

so, w = v/r

let m is the mass and r is radius of the sphere.
now apply conservation of energy

initial potential energy at height h = sum of potential and kinetic energy at the top of the loop

PEi = PEf + KEf

PEi = PEf + KE_linear + KE_rotational

m*g*h = m*g*(2*R) + (1/2)*m*v^2 + (1/2)*I*w^2

m*g*h = 2*m*g*R + (1/2)*m*v^2 + (1/2)*(2/5)*m*r^2*w^2

m*g*h = 2*m*g*R + (1/2)*m*v^2 + (1/5)*m*(r*w)^2

m*g*h = 2*m*g*R + (1/2)*m*v^2 + (1/5)*m*v^2

m*g*h = 2*m*g*R + (7/10)*m*v^2

m*g*h = 2*m*g*R + (7/10)*m*(5*g*R)

m*g*h = m*g*R + 3.5*m*g*R

m*g*h = 4.5*m*g*R

h = 4.5*R (or) (9/2)*R <<<<<<<<<-----------------Answer