Question on coriolis effect with drag force

AI Thread Summary
The discussion centers on the trajectory of a small floating object influenced by drag and Coriolis forces. The drag force is defined as F = -αv, where α is a constant coefficient related to the drag force as a function of velocity. There is debate over whether the drag should be proportional to velocity or the square of velocity, with some suggesting that a low Reynolds number scenario could justify the linear relationship. The Coriolis force acts perpendicularly to the object's speed, complicating the trajectory analysis. Ultimately, the conversation emphasizes the need for a kinematic analysis that considers all acting forces and the effects of Earth's rotation.
ystch
Messages
1
Reaction score
0
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 ?
 
Physics news on Phys.org
What is "α"?
 
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'
 
256bits said:
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.

willem2 said:
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.
 
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 ).
 
256bits said:
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.
 
Aero_UoP said:
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.

Or it's on earth, in a flow condition dominated by viscous rather than inertial forces (a flow with a very low reynolds number).
 
cjl said:
Or it's on earth, in a flow condition dominated by viscous rather than inertial forces (a flow with a very low reynolds number).

Creeping flow... you're right, it slipped my mind.
I haven't dealt with such a flow for like a century :p lol
 
Back
Top