# Question on coriolis effect with drag force

I really need help with this question.

A small floating object initially moves with velocity v on the surface of a liquid at latitude λ. The drag force due to liquid is F=-αv. Find the shape of the trajectory of the object due to drag and Coriolis forces. What will be the trajectory if α=0 ?

What is "α"?

256bits
Gold Member
a would be a constant coefficient in this equation that relates drag force as a function of velocity.

the equation is of similar from as friction where F = u N.

The coriolis force is perpendicular to the speed, so the drag isn't influenced by the angle that the object is moving. You can solve a differential equation for the maginuted of the speed first: x'' = -ax'

a would be a constant coefficient in this equation that relates drag force as a function of velocity.

the equation is of similar from as friction where F = u N.

Then α has to be 1/2*ρ*Cd*S*u where ρ=density, Cd=coefficient of drag, S=the wetted area and u the velocity because drag is a function of the square of the velocity.

The coriolis force is perpendicular to the speed, so the drag isn't influenced by the angle that the object is moving. You can solve a differential equation for the maginuted of the speed first: x'' = -ax'

The drag is indeed influenced by the angle that the object is moving. If it sideslips drag is higher than if it was moving with 0 angle from the longitudinal axis.

ystch,
If I were you, I would perform the kinematic analysis. Consider all forces acting on the object and take under consideration earth's rotation as well.
α=0 means probably that you ignore the drag so you do the analysis once more neglecting drag.

256bits
Gold Member
Then α has to be 1/2*ρ*Cd*S*u where ρ=density, Cd=coefficient of drag, S=the wetted area and u the velocity because drag is a function of the square of the velocity.

The equation for the problem does not say that.
It says that the drag force has a direct relationship with velocity ( not velocity squared ).

The equation for the problem does not say that.
It says that the drag force has a direct relationship with velocity ( not velocity squared ).

I know. That's why I say that the equation given is wrong, unless the problem is not on earth but somewhere else that drad could depend on the velocity and not on the square of the velocity ;) Under this assumption, he can proceed the solving procedure with the given equation.

cjl