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(II) A helicopter rotor blade can be considered a long thin rod, as shown in Fig. $55 .(a)$ If each of the three rotor heli- copter blades is 3.75 $\mathrm{m}$ long and has a mass of 135 $\mathrm{kg}$, calculate the moment of inertia of the three rotor blades about the axis of rotation. (b) How much torque must the motor apply to bring the blades from rest up to a speed of 5.0 $\mathrm{rev} / \mathrm{s}$ in 8.0 $\mathrm{s} ?$

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(a) $I_{to t}=1898 \mathrm{~kg} \cdot \mathrm{m}^{2}$(b) 7500 $\mathrm{m.N}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Christine N.

August 11, 2020

Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

Lectures

02:21

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.

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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(II) A helicopter rotor bl…

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A helicopter rotor blade c…

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(II) Let us treat a helico…

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03:34

helicopter rotor blade can…

01:48

04:31

Part A: A helicopter rotor…

03:15

Let us treat a helicopter …

02:14

Helicopter Blades Each of…

01:25

Each of three helicopter r…

01:36

A helicopter has two blade…

01:20

03:54

A uniform helicopter rotor…

17:40

04:06

The rotor (flywheel) of a …

02:25

The lightweight turbine co…

So here for part A. We know the moment of inertia, of a thin rod rotating about its end, we can say the moment of inertia. Book the Rod again. This would be rotating about its end with equal 1/3 ml square. It feels rotating through the center of Mass. That would be a completely different moment of inertia. We have them three blades and so the total moments of inertia, with the equaling three times 1/3 ml square, which will simply be equal to M L Square and we can solve. So this would be 135 kilograms multiplied by 3.75 meters, quantity squared. And so the total moment of inertia four part a would be 1898 kilograms meter squared. This would be our answer for part a four part B. Then we need to find the torque required, and we can say for part B, the torque required would be equaling to the total moment of inertia multiplied by the angular acceleration Alfa. This would be the total moment of inertia and according to the definition of the change in, uh according to the definition of the angular acceleration. This would be equaling two Delta Omega, divided by Delta T or the change in the angular velocity with respect the time. And so this would be 1898 kilograms meters squared and then multiplied by 5.0 revolutions for a second, multiplied by two pi radiance for every revolution. And then this would be divided by the time of 8.0 seconds, and we find that the torque required would be equaling 7500 Newton meters. This would be approximately 7500 Newton meters rounded to two significant figures. That is the end of the solution. Thank you for

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