What is the Force Exerted by a Falling Rod on a Hinged Support?

[Chestermiller]

For drag flow past a circular cylinder, the drag coefficient is nearly constant on the order of about 1.0 for Reynolds numbers on the order of about 1000 to 100,000. So, the drag force per unit length on the rod is $$f=\frac{1}{2}\rho v^2 D$$where D is the diameter and v is the velocity across the rod (the component of velocity normal to the rod). So $$f=k(\omega r)^2=k'r^2$$where r is the distance from the hinge. So, to a good approximation, Haruspex's exponent is 2. The moment of the drag force is $$M=FL\cos{\theta}=\int_0^L{k'r^3dr}=k'\frac{L^4}{4}$$So, $$k'=\frac{4F}{L^3}\cos{\theta}$$This means that the local normal force per unit length exerted by the fluid on the rod is:
$$f=\frac{4Fr^2}{L^3}\cos{\theta}$$So, what is the normal force that the fluid exerts on the rod?

 

[haruspex]

hahaha yea I know. And I agree. So how would we proceed here? We balance these two torques but how will this help us resolve that force?

When you have found the total normal force on the rod from the fluid (as per Chet's post) you can similarly find the total torque from the fluid.
This gives you effectively three balance equations, two being the two dimensional balance of linear forces, and one from the balance of torques. That means you can find both components of the hinge reaction even without knowing the A factor in the force from the fluid.
As Chet says, n=2 is reasonable.

 

[Chestermiller]

When you have found the total normal force on the rod from the fluid (as per Chet's post) you can similarly find the total torque from the fluid.
This gives you effectively three balance equations, two being the two dimensional balance of linear forces, and one from the balance of torques. That means you can find both components of the honge reaction even without knowing the A factor in the force from the fluid.
As Chet says, n=2 is reasonable.

I used the total torque to get the value of the constant k'. All he needs to do now is determine the total normal component of the fluid force.

 

[haruspex]

I used the total torque to get the value of the constant k'. All he needs to do now is determine the total normal component of the fluid force.

Ah yes, I didn't read your post in enough detail. I think it's actually a bit simpler my way, to write out all the equations assuming the drag is at distance r from the hinge is Ar². The A's cancel without ever having to determine it; in particular, the dependency on the angle.

 

[member 428835]

This means that the local normal force per unit length exerted by the fluid on the rod is:
$$f=\frac{4Fr^2}{L^3}\cos{\theta}$$So, what is the normal force that the fluid exerts on the rod?

$$F_n=\frac{4Fr^3}{3L^3}\cos{\theta}$$ Thank you both for the input!

 

[Chestermiller]

$$F_n=\frac{4Fr^3}{3L^3}\cos{\theta}$$ Thank you both for the input!

Good, except lose the r^3 and the L^3

 

[kuruman]

The moment of the drag force is $M=FL\cos{\theta}=\int_0^L{k'r^3dr}=k'\frac{L^4}{4}$

Question: Why is the upper limit of integration L? The drawing attached to post #16 shows the fluid extending only partially along the full length of the rod. Or is it just a drawing?

 

[Chestermiller]

Question: Why is the upper limit of integration L? The drawing attached to post #16 shows the fluid extending only partially along the full length of the rod. Or is it just a drawing?

That's not how I interpreted the diagram. I don't know what the correct interpretation is. Josh?

 

[member 428835]

Yea, my drawing is imperfect. Fluid is coming out of the entire triangular region.

 

[TSny]

The initial question posed was asking what force the hinge exerts on the ground.

There is no mention of the ground in the OP. Have you stated the problem word-for-word as given?

Not that I think I can contribute anything here, but I just wanted to understand the statement of the problem. I can't make any sense out of it whatsoever.