Question

A glass filled with water is whirled in a verticle circle of radius `R` what can be the minimum speed at the top of the path if water does not fall out from glass? If the glass moves with this speed, find the normal contact force the glass exerts on water at the lowest point of the path?

Answer 1

At the highest point :
`N + mg = (mv^(2))/(R ) implies N = (mv^(2))/(R ) - mg`
If water does not fall, then
`N ge 0`
`(mv^(2))/(R) - mg ge 0 implies v ge sqrt(gR)`
`v_(min) = sqrt(gR)`
At the lower point:
`N_(0) - mg = (mv^(2))/(R ) = mg`
`N_(0) = 2 mg`