MIT16_07F09_Lec12

MIT16_07F09_Lec12 - S. Widnall, J. Peraire 16.07 Dynamics...

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Unformatted text preview: S. Widnall, J. Peraire 16.07 Dynamics Fall 2008 Version 2.0 Lecture L12- Work and Energy So far we have used Newtons second law F = m a to establish the instantaneous relation between the sum of the forces acting on a particle and the acceleration of that particle. Once the acceleration is known, the velocity (or position) is obtained by integrating the expression of the acceleration (or velocity). Newtons law and the conservation of momentum give us a vector description of the motion of a particle in three dimensions There are two situations in which the cumulative effects of unbalanced forces acting on a particle are of interest to us. These involve: a) forces acting along the trajectory. In this case, integration of the forces with respect to the displacement leads to the principle of work and energy . b) forces acting over a time interval. In this case, integration of the forces with respect to the time leads to the principle of impulse and momentum as discussed in Lecture 9. It turns out that in many situations, these integrations can be carried beforehand to produce equations that relate the velocities at the initial and final integration points. In this way, the velocity can be obtained directly, thus making it unnecessary to solve for the acceleration. We shall see that these integrated forms of the equations of motion are very useful in the practical solution of dynamics problems. Mechanical Work Consider a force F acting on a particle that moves along a path. Let r be the position of the particle measured relative to the origin O . The work done by the force F when the particle moves an infinitesimal amount d r is defined as dW = F d r . (1) 1 That is, the work done by the force F , over an infinitesimal displacement d r , is the scalar product of F and d r . It follows that the work is a scalar quantity. Using the definition of the scalar product, we have that dW = F d r = Fds cos , where ds is the modulus of d r , and is the angle between F and d r . Since d r is parallel to the tangent vector to the path, e t , (i.e. d r = ds e t ), we have that F e t = F t . Thus, dW = F t ds , (2) which implies that only the tangential component of the force does work. During a finite increment in which the particle moves from position r 1 to position r 2 , the total work done by F is r 2 s 2 W 12 = F d r = F t ds . (3) r 1 s 1 Here, s 1 and s 2 are the path coordinates corresponding to r 1 and r 2 . Note Units of Work In the international system, SI, the unit of work is the Joule (J). We have that 1 J = 1 N m. In the English system the unit of work is the ft-lb. We note that the units of work and moment are the same. It is customary to use ft-lb for work and lb-ft for moments to avoid confusion....
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This note was uploaded on 11/06/2011 for the course AERO 112 taught by Professor Widnall during the Fall '09 term at MIT.

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MIT16_07F09_Lec12 - S. Widnall, J. Peraire 16.07 Dynamics...

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