Objects that rotate in air or water experience a torque due to drag. With quadratic drag, a drag torque that's negligible at low rpm quickly becomes significant as the rpm increases. Consider a square bar with cross section a x a and length L. It is rotating on an axle through its center at angular velocity ω in a fluid of density p. Assume that the drag coefficient C𝒹 is constant along the length of the bar. Find an expression for the magnitude of the drag torque on the bar.
Hint: Begin by considering the drag force on a small piece of the bar of length dr at distance r from the axle.

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Accepted Answer

Hi everyone. In this practice problem, you will be asked to derive an expression for a rod drag torque magnitude. It is common for friction to create that torque that accelerates rotating objects. When objects rotate in fluids, it creates drag force that has the same consequences as friction drag has a characteristic nature where it is nearly negligible at low rotational speed and increases very fast at high rotational speeds. Using a circular rot of diameter D length Y rotating at the angular velocity omega in a fluid whose density is row taking the drag coefficient to be C and assuming it is constant at all points on the rot for the rotation on an axle to one of its ends and perpendicular to the length will be asked to derive an expression of the rot's drag to magnitude and consider the drag force on an element dr located at a distance R from the rotational axis. The option listing the different expressions for the rot direct toque magnitude. So in this problem, we'll consider the direct to be quadratic. We also assume that the coefficient of Dr which is given in the problem statement to be C is going to be constant along the rectangular rot. We can express the direct force at a high quadratic num number as fre which if we recall F track will be equal to half multiplied by uh row multiplied by C multiplied by a three square. So let's consider a small element of the rectangular rot. The R located at a distance R from the axis of the rot. Such a small speed of the rectangular rot experiences speed through space which is given by V equals to omega multiplied by R. If you recall to can be represented with Tau, which will be equal to R multiplied by F. In this case, to direct will be equal to R multiplied by F direct. Which if we substitute our F track equation, this equation for Tau drag will then be equal to half multiplied by C multiplied by row multiplied by A P squared and R. So in this case, we also know that V will be equal to omega multiplied by R. So we can substitute RV squared value here in order for us to obtain an equation with omega and R. So Tau direct substituting the V will then be equal to half multiplied by C multiplied by a row. A omega squared R squared multiplied by R just like. So, so a cylinder will have circular and rectangular cross sections and the curve section will have a rectangular cross section considering the rotational axis, the area of the cross section affected by drag is a rectangle. So the A that is given in the top drag here is going to be equal to the area of a rectangle, which is going to be equal to diameter multiplied by a length. So in this case, that will be D multiplied by Y or D multiplied by R. So now substituting our area equation that we just have, which is D multiplied by R into the drag torque. We will then be able to obtain the formula that D or Tau drag will then be equal to R. I'm just gonna put uh pull the R here in front multiplied by half C row, Dr Omega squared, R squared and then sort of lumping the RS together on the right side, we will have R multiplied by half C row D Omega square R cube. So that will be the torque uh direct torque formula. And now we want to consider the contribution from a differential segment which is uh D tau direct will then be equal to because um Tau has the formula of R multiplied by F and Tau drag will have the formula of R multiplied by F Dr. So based on chain rule, the D multiplied by T direct uh will be equal to R multiplied by DF Dr plus F multiplied by Dr. And then substituting the values that we are the expressions that we have the will then be equal to R multiplied by DF drag will be equal to three divided by two multiplied by C row D omega squared R squared multiplied by Dr plus health, which is uh DF here is going to be F track which is going to be half multiplied by C row D omega squared R cube multiplied by Dr and then simplifying things. We will then be able to obtain three divided by two multiplied by C row D omega squared R cube multiplied by Dr plus half C row D Omega squared RQ multiplied by Dr. And then calculating or combining everything we will then get two C row D omega squared R cube. Dr just like. So, so now we'll consider the contribution to drag from elements on the rot. So a total drag is the sum of the contribution from all the elements. And the implication is that different segments of the rectangular cross section will contribute to the total drag torque. So we'll integrate from R equals to zero to the axis of R equals to Y at the farthest end. So Tau direct will then be equal to our formula that we just obtained two row D omega squared. And then we are integrating from, from R equals to zero to R equals to Y. So the integral from R equals to zero to R equals to Y of RQ multiplied by the R. And then we want to recall that the integration principle that we have the integral of RQDR will be equal to one divided by four, multiplied by R to the power of four plus C obviously. But since it is uh bounded in the goal, so we just have to um evaluate them. So this will then be equal to two row D omega squared. And this uh are just the constant that we have. So two row D omega squared multiplied by R to the power of four divided by four, evaluated from zero to Y. So then the Tau direct will just be equal to half multiplied by C row D omega squared Y to the power of four. So the interpretation of our power that we have here, the expression actually means that drag torque will increase quadratic with omega squared and even faster with Y to the power of four. So if we move our axis to stay in the middle of the rod and we have more arms around the axis, we'll form an expression for each arm integrating that and add the integrated expression to form the total drag torque. So there we have our final expression. So drag will be equal to C row D omega squared Y to the power of four divided by two. And that will correspond to option A and our answer choices. So option A will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.

Verified Solution

Key Concepts

Torque

Drag Force

Quadratic Drag