# Integrating Torque With Force as a Function of Radius

1. Apr 5, 2009

### ztluhcs

1. The problem statement, all variables and given/known data

In this problem, I am asked to find the total torque acting on a device consisting of two open half-cylinders connected to a shaft rotating in a fluid. To clarify, they are half-cylinders in the sense that they have been cut in half in the lengthwise direction and are oriented to generate drag. In the (very rough) picture below the spanwise elements are the two half cylinders (===) and the vertical part is the shaft (++).

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2. Relevant equations
T = r x F

F = 1/2 * rho * S_ref * C_D * V^2 = 1/2 * rho * S_ref * C_D * (omega * r)^2

where rho, S_ref, and C_D, the angular velocity (omega), the diameter, and the radius of the cylinder are all known

3. The attempt at a solution

As per my instructor's advice, the force on one cylinder is given in the equation above. Then a differential element of torque would be written as:

dT = 1/2 * rho * S_ref * C_D * (omega * r)^2 * r*dr

integrating over 0 to R should then give the total torque, which is (for both cylinders):

T = 1/4 * C_D * rho * D * omega^2 * R^5

The problem with this solution is that the units are kg*m^3/s^2 and the units of torque are kg*m^2/s^2, so there is some dimensional inconsistency. Instead I was thinking that differential radius in the cross-product should be just dr, but looking back I'm also not sure if it is necessary to take some differential element of the force (which of course itself has velocity as a function of radius. I'm just having some issues applying physical intuition to this fairly complicated situation since the force changes as radius changes.

Thanks for any help,
Ian

2. Apr 6, 2009

### Redbelly98

Staff Emeritus
I think you need to come up with an expression for dF first, then multiply it by r to get dT.

Something in the F equation must represent area, and would thus would contain the necessary dr term when you write dF.

3. Apr 6, 2009

### ztluhcs

Well, velocity is a function of radius, so you can write F = constants * r^2 so then

dF = 2*constants*r*dr

This would give:

dT = dF x r

dT = 2 * constants * r^2 * dr

T = 2/3 * constants * R^3

which is different than either of my other proposed solutions! Does this seem correct?

4. Apr 6, 2009

### Redbelly98

Staff Emeritus
No, dF should be proportional to r2.

Think of a differential area element dSref, which has a drag force acting on it. Express dSref in terms of dr.

p.s. Welcome to Physics Forums!