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Ch 09: Rotation of Rigid Bodies

Chapter 9, Problem 9

An airplane propeller is 2.08 m in length (from tip to tip) with mass 117 kg and is rotating at 2400 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

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Hello everyone. So in this problem, a wind turbine blade rotates at 10 revolutions per minute, about its central axis and the length of the blade is five m and its mass is 20 kg. Blade can be considered like a thin rod first. We want to determine the blades, rotational kinetic energy and seconds to practical constraints. It will be more convenient to divide the weight of the blade by two. What would be the angular speed needed in order to maintain the same blade size and same Connecticut so we can split this problem into two first. We want to find the kinetic or the rotational kinetic energy. So we're given that the blade can be considered like a thin rod. We recall that the kinetic energy for a rotating object is one half. I, omega squared. We also recall that for a full rod. The equation for i It's just 12. And elsewhere from the problem statement, we know that it rotates 10 revolutions per minute. So we are given omega and now we can solve for the kinetic energy plug in the values This is a two, 1 12 Where M is 20 kg is five m square. And for omega Were given 10 revolutions a minute but we want to convert this to radiance for a second to get the correct units to match up. So we know that one revolution is too high radiant and that There are 60 seconds. This whole term is split. You get the kinetic rotational energy, It's 22. second part. We're told We want to divide the weight of the blade by two To the ratio between M- one M 2. You know that M two is going to be one half and one, So I have one over .5 and one these two cancel out, let me know that 1/1 half is just too. So to incorporate this M1 over M2 ratio. We can take the ratio between the kinetic energies because we know there needs to be the same kinetic energy between the two scenarios. So we sold for K one in the first part, Divided by K two. In the second part it will just be one a half. I won omega one squid over one a half, I to omega to squid since it is the same kinetic energy. This ratio will be equal to And we can also serve in I one and I two. So we get that one for K one, it is one a half And then I one is going to be 12 M one L one square. And we also carry the omega one squared term. And on the bottom we have 1/2 times I two which is 1 12 M two. L two squared. We also carry the Omega two square. We can see that there's a lot of terms to cancel out one half to cancel out 1 12 to cancel out were given that the blade must maintain its size so the L one and L two are equal. So finally we have the K1 or KE two which is the same, so it is equal to one. This ratio simply equal to M one over M two times omega one squared over omega two squared from the first part of the problem. Semen, We know what Omega one is and We also calculated the ratio and one over M two. So now we can rearrange this equation to find Omega two. I got two squared is equal to 9, 1 over and two Times Omega one Squid. And now we just take the square root Mega two is equal to the square root 12 Times Omega one Squared. So finally I did too square roots two Times Omega one square, Which is 10 rpm. This gives us an answer of 14.1 revolutions. So the answer, the first part is 22.8 jewels in the second part, 14.1 for a minute. So this gives us an answer of choice D. Of themselves have a great day
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