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An airplane propeller is 2.08 m in length (from tip to tip) with mass 117 kg and is rotating at 2400 rpm v 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?

a) $1.332 \times 10^{6} \mathrm{J}$b) 2771 $\mathrm{rev} / \mathrm{min}$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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An airplane propeller is 2…

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The first step to solving this problem is to figure out what the moment of inertia of the Rod is equal to in this case of the propeller. And by looking at table nine, too, we see that has the form one twelve 12 times the mass times that length squared. And then the problem gives us both of these values so we can solve to get a value of one point. Sorry, wrong thing here to get a value of 42.2 kilograms meter squared. Now. They also give us being more velocity as 24 100 rpm's. So we need to convert this in Iranians for second, like converting rotation so radiant and menaces seconds. When you do that, we get a value of 251 radiance per second. And this is the value that we need to plug in into the formula's because it's in the S i u nits. So the kinetic energy has the form. 1/2 time's a moment, inertia times and your velocity squared. We found both the mill owners here and angular velocity here, and so my employers in we get 1.33 times 10 to the sixth rules. And so that is the answer to party in part B. Let's write out what the initial kinetic energy is before the change is the more inertia, which is 1/12 M one that once where Times the angular velocity. So it's Omega one square. There's actually a 1/2 hour in front, too. And so it's 1/2 I, which is this whole thing Times Omega squared now. Okay, too, which is after the change. It's still going to have the 1/2 and it's also going to still havin a moment inertia. In this case, I'm going to switch the variables to have a sub script of two. And so I have into l two squared and then Omega two square. And so this is the kinetic energy before the change. And this is the connect energy. After the change, we know that l one equals l, too. So the length of the propeller doesn't change through the change here. We also know the connect energy doesn't change. So these two are constants. What does changes into an omega, too? We know how the mass changes and we're trying to sell for how the angular velocity changes. So because the kinetic energies are equivalent, I'm going to equate them together. Since Katie was equal. Okay, one if came one is equal to this. In case one is also a little of this, then these two things are equal to each other. So this means that one twelve 12 or 1/2 times one twelve 12 times in one ln squared omega one square is equal to 1/2 times one twelve 12. I am too. Lt squared a maid too square. Now you'll see the one half's cancel the one twelves cancel. And since l one is equal to hell too, the length factors cancel and so are left of someone Omega One square is equal to 12 Omega two squared. And we want to solve this for Omega too. When I do that, you don't make it too is equal to omega one times the square root of M one over into. And we know that the mass after the change is equal to three. Force the mass from before the change. So am to his equal 2.75 in one. And so you can cancel them ones like that. Now we know Omega one in Rpm's is equal to 24 100. And so we have everything we need. And when we saw Sal, we get will make it, too, is equal to 2,770 rpm's. And so it's increased somewhat dramatically by 370 rpm's. That's final answer.

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