Physics: A mass falling off a Globe?

Question

A large globe, with a radius of about 5m , was built in Italy between 1982 and 1987. Imagine that such a globe has a radius R and a frictionless surface. A small block of mass m slides starts from rest at the very top of the globe and slides along the surface of the globe. The block leaves the surface of the globe when it reaches a height h above the ground.

Using Newton's 2nd law, find Vcrit, the speed of the block at the critical moment when the block leaves the surface of the globe.
1. Assume that the height at which the block leaves the surface of the globe is hcrit.
Express the speed in terms of R, hcrit, and g, the magnitude of the accleration due to gravity. Do not use in your answer theta.

2.Use the law of conservation of energy to find vcrit . This will give you a difference expression for than you found in the previous part.
Express in terms of R, h, and g.

3.Find the height from the ground at which the block leaves the surface of the globe. Expressin termsR

Answer 1

We're interested in only the upper hemisphere, so for simplicity we'll define h and hcrit as the height above the sphere's equator. (Later you will have to add r to get the actual hcrit.) As the block slides, it requires v^2/r centripetal acceleration to keep it on the sphere. This is provided by gravity, which in the radial direction is g*h/r. At hcrit, v^2/r = g*h/r, so vcrit^2 = g*hcrit (ans. 1). The velocity at any h is derived from conservation of energy. The potential energy converted to kinetic energy is m*g*(r-h) and the kinetic energy is .5*m*v^2, so at hcrit, vcrit^2 = 2*g*(r-hcrit) (ans. 2). Now we have two equations for vcrit^2 involving hcrit so we can solve for hcrit: vcrit^2 = g*hcrit = 2*g*(r-hcrit), thus hcrit = 2*r/3. Then vcrit can be defined independently of hcrit: vcrit = sqrt(2*g*r/3). As for (3), the height above the ground is r + hcrit, or 5*r/3.

Answer 2

v_l = sqrt g(h_l-R) for the first one
v_l = sqrt 2g (2R-h_l) second one
h_l = 5R/3 for third one