Max speed and displacement of particle on rotating slope with Coriolis force

[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]

thank you!The maximum speed acquired by the particle can be found by setting the net force on the particle equal to zero. In this case, the net force is the gravitational force down the slope minus the Coriolis force. The Coriolis force is equal to 2mωv, where v is the velocity of the particle. The gravitational force down the slope is equal to mgsinθ, where g is the gravitational acceleration and θ is the angle formed by the plane with the horizontal. Setting the net force equal to zero and solving for v gives:v = gsinθ/2ω The maximum downhill displacement can be found by integrating the velocity equation above (assuming the particle starts from rest at x = 0) to get the following expression for the position of the particle x:x = gsinθ/(4ω^2) Therefore, the maximum downhill displacement of the particle is gsinθ/(4ω^2).