Question

A skier stands on the top of a giant frictionless snowball with radius R:She is initially at rest, but a slight perturbation starts her sliding off to the side. (a) At what angle with respect to the vertical will she lose contact with the snow? (b) What is the velocity at this point if R = 50 m, and how fast will she be traveling when she hits the ground? The snowball is fixed in place, and doesn't roll.

Answer 1

a] Let the angle be theta.

Centripetal force = mv^2/R

mg cos theta = mv^2/R.......equation 1

By energy conservation, 0.5mv^2 = mgR [1 - cos theta]

mv^2/R = 2mg [1 - cos theta]

putting this value in equation 1,

mg cos theta = 2mg [1 - cos theta]

cos theta = 2 - 2 cos theta

cos theta = 2/3

theta = arccos (2/3) = 48.2 degree

b]  mv^2/R = 2mg [1 - cos theta]

when she leaves v = sqrt(2Rg (1-cos theta)) = sqrt(2*50*9.8*1/3) = 18.07 m/s answer

when she hits ground, v = sqrt(2*50*9.8) = 31.3 m/s answer