A radio transmission tower has a mass of 76 kg and is 12 m hig...

Question

A radio transmission tower has a mass of 76 kg and is 12 m high. The tower is anchored to the ground by a flexible joint at its base, but it is secured by three cables 120° apart (Fig. 11–52). In an analysis of a potential failure, a mechanical engineer needs to determine the behavior of the tower if one of the cables breaks. The tower would fall away from the broken cable, rotating about its base. Determine the speed of the top of the tower as a function of the rotation angle θ. Start your analysis with the rotational dynamics equation of motion d $\vec{L}$ /dt = ${\vec{τ_{e x t}}}_{}$ . Approximate the tower as a tall thin rod.

Diagram of a radio transmission tower, showing its rotation angle θ, forces, and motion equations for analysis.

Accepted Answer

Welcome back. Everyone in this problem. A meteorological mast used for measuring wind speed and atmospheric data has a mass of 120 kg and stands 25 m tall. This mast is supported by three guy wires each attached at 120 degree intervals around the mast base to assess the failure modes. An analysis is conducted on the scenario where one of the guy wires breaks the mast would then rotate about its base opposite the broken wire determine the tip speed of the mast as a function of its tilt angle theta relative to the vertical by applying the rotational dynamics equation delta L with respect to time equals external torque. As a hint, we can approximate the tour as a tall thin rod. For our answer choices. A says that the speed is 15 multiplied by the square root of one minus K. The meters per second. B says it's 27 multiplied by the square root of one minus cas theta meters per second. C says it's 27 multiplied by the square root of two multiplied by one minus costs the meters per second. And D says it's 54 multiplied by the square root of two multiplied by one minus costs the meters per second. Now, in this problem, essentially, all we are trying to do is to figure out the tip speed as a function of its tilt angle data. OK. And we are already told that we are applying the rotational dynamics equation which says the external torque is equal to the change in angular momentum over time. Now we assume that the tensions in the two intact guide wires drop to zero immediately. Thus, having no impact on the motion, the forces acting on the meteorological must include the forces at the base joint and the weight, the rotation axis is located at the ground attachment point. So we can say that it is located here. And since this axis is stationary in an inertial frame, we can apply our equation that is some of the torque caused a change in angular momentum over time. Now, what do we know about torque that can help us in this problem? Well, we know that our torque, OK is going to be equal to our force multiplied by the perpendicular distance. So in this scenario, then the force we are looking at the force applied is the weight of the mass acting vertically downwards and perpendicular to that weight is the horizontal component of half of the mass length. So that means then that the sum of the ta is going to be equal to MG multiplied by A half of L sine theta. So all of that is going to be equal to our change in angular momentum over time. But again, what do we know about angular momentum? Well, we also know that our angular momentum L equals our moment of inertia multiplied by our angular velocity. OK. So that means our change in angular momentum DL is going to be equal to D I omega. And in this case, since our mass is falling, then where its angular velocity is changing. So we could rewrite this as I multiplied by delta omega. So if we apply that idea, then OK, then that means for what we already have here in red, we could then say that a half of MGL sine theta is going to be equal to I multiplied by D Omega with respect to time. In other words, a change in angular velocity with respect to time. Now, remember we said that we approximated the Torah as a tall thin rod and the moment of inertia for a tall thin rod is half of its mass multiplied by its length squared. So that means then that we could say a half of MGL sine theta is going to be equal to one third of ML squared D OMEGA with respect to time. I'm trying to remember if I said one third ML squared. I think I did before. I hope I did. But yes, it's one third ML squared. Now we can take this a step further. OK. Let's try to simplify. So notice we can cancel M from both sides and we can cancel L from our left hand side and L from our right hand side from L squared to leave us with just L. So that means that a half of G sine theta is going to be equal to a third of L multiplied by A or change in angular velocity over time. But again, we could write our change in angular velocity as our change in angular velocity with respect to our tilt angle multiplied by a change in tilt angle with respect to time. Now, how can that help us? Well, our change in tilt angle or our change in angular displacement with respect to time is defined as our angular velocity. So we could replace this expression then OK with Omega which leaves us then with one third L delta omega with respect to theta multiplied by Omega. Now, if we try to isolate our angular velocity on one side and our theta terms on the other side, OK, then we're going to end up with three G divided by two L sine theta delta theta being equal to Omega delta omega. And now at this point, OK, we can integrate both sides, we can integrate the left hand side with respect to theta and we can integrate the right hand side with respect to Omega. OK. Now, notice on the left hand side, we have some constants here. So we can write this as three G divided by two L multiplied by the integral of sine theta delta theta being equal to the integral of Omega delta omega. OK. And when we do integrate, then we find that we end up with three G divided by two L multiplied by one minus cost theta being equal to omega squared divided by two because that's the integral of omega. No. OK. We also know well here at this point, sorry, we can start to solve for Omega. OK. So when we do that, then notice we can cancel two from both sides. And now we can say that three G divided by L multiplied by one minus cost, theta equals omega squared. Now remember we want to express our speed or speed as a function of uh of, of its tilt angle, not our angular velocity. So we also know we also know that what we can use to help us here. We also know that V equals L multiplied by Omega. So Omega equals V divided by L. Now, that's pretty helpful because now we've been able to incorporate our speed into our expression. And if we rewrite it with omega squared on the left, then that's going to be V divided by L all squared being equal to three G divided by L multiplied by one minus cos theta no, if we square both terms, that's going to be, let me get rid of our bracket here. So that's gonna be V squared divided by L squared. OK. And now, OK, we can multiply both sides by L squared. So V squared equals three G divided by L multiplied by L squared multiplied by one minus theta. We can cancel L from our denominator. So we can get V squared to be equal to three GL multiplied by one minus K theta. And now that means V is going to be equal to the square root of three GL multiplied by one minus cos theta. Now, at this point, we can find the square root of both terms. So as the square root of three gl multiplied by the square root of one minus cost theta, I mean this is a term on its own here. And the reason why we did that is because we know what G and L are, we can take G the acceleration due to gravity to be 9.8 m per second squared. And we know that L, the length of the mast is 25 m. So that means then that V is going to be equal to the square root of three multiplied by 9.8 m per second squared, multiplied by 25 m multiplied by the square root of one minus cos theta. Now, when we find the square root of this expression, OK, we get it to be 27.11 multiplied by one minus cost the, and if we were to round our term here to the nearest or to two significant figures. Then we'll get V or speed to be 27 multiplied by the square root of one minus cos theta meters per second. Thus, we can use that expression as the tip spirit of the mast as a function of its tilt angle data. If we go back to our answer choices that tells us that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Rotational Dynamics

Torque

Angular Velocity and Acceleration

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