Solving Dynamics Problem: Coriolis & Gravitational Forces on Rotating System

[lovelyrita24]

I have a problem involving the coriolis and gravitational forces on a rotating coordinate system:

A free particle of mass m is release from a state of rest on a rotating, sloping, rigid plane. The angular rotation rate about a vertical axis is omega and the angle formed by the plane with the horizontal is theta. Friction and centrifugal forces are negligible. What is the maximum speed acquired by the particle, and what is its maximum downhill displacement?

- I am pretty sure the answers are derived symbolically by determining when the coriolis force comes into balance with the component of the gravitational force down the slope.

Any insight would be much appreciated...

 

[member 553083]

The maximum speed acquired by the particle can be determined by setting the net force acting on it equal to zero. This would give us the equation:$m \omega^2 r \sin(\theta) - mg \cos(\theta) = 0$where $r$ is the distance of the particle from the axis of rotation. Solving for the maximum speed, we get:$v_{max} = \frac{mg \cos(\theta)}{m \omega \sin(\theta)}$The maximum downhill displacement can be determined by noting that the total energy of the system is conserved. The initial energy is zero, so the final energy must also be zero. We can calculate the final energy as:$E_{total} = \frac{1}{2} m v^2 - m g y \cos(\theta)$where $y$ is the displacement of the particle down the slope. Setting this equal to zero and solving for $y$, we get:$y_{max} = \frac{v^2}{2 g \cos(\theta)}$Substituting in our expression for $v_{max}$, we get:$y_{max} = \frac{\left (\frac{mg \cos(\theta)}{m \omega \sin(\theta)} \right )^2}{2 g \cos(\theta)} = \frac{g}{\omega^2 \sin^2(\theta)}$