Question

A boy is sat at a semisphere with a ray of 18.4 m. He starts sliding. On what height will the boy loose contact with the sphere? There is no friction.

Answer 1

12.3 m

Explanation:

Key concepts needed for the solution:
- Centripetal force as a net force `F_c= mv^2/r`
- Conservation of energy `1/2mv^2+mgh =` constant

r = radius of the sphere
`theta` = angle of a ray with respect to the horizontal
`h = rsintheta` = high at which the boy loose contact with the sphere.

When the boy is sliding down the hemisphere, he experiences a centripetal force:

`F_c= mv^2/r`

There are two forces contribute the centripetal (net) force, the normal force and gravitation pull. Note that the boy's motion at any instant on the sphere is tangential to the hemispheric arc. It can be shown by geometry that the force pushing radially on to the sphere due to the boy's weight is `mgsintheta`. This force supplies both the normal force and the centripetal force.

`therefore F_c=mgsintheta-N= mv^2/r`

At the moment the boy loss contact with the sphere, N=0.
`rArr mgsintheta +0 = mv^2/r`
`rArr mv^2 = mgrsintheta=mgh`

Friction free means that we can apply conservation of energy to this scenario. And together with the equation above, we get:

`cancel(mg)r=1/2mv^2+mgh=1/2mgh+mgh=3/2cancel(mg)h`

`rArr 3/2h =r`

`rArr h= 2/3r =2/3(18.4m)= 12.3m`