Work1 - To compute the work we calculate the component of...

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Work Given our definition of work as W = Fs , can we generate an expression for work done on a rotational system? To derive our expression we begin by taking the simplest case: when the force applied to a particle in rotational motion is perpendicular to the radius of the particle. In this orientation, the force applied is parallel to the displacement of the particle, and would exert the maximum work. Given this situation the work done is simply W = Fs , where s is the arc length that the force acts through in a given period of time. Recall, however, that arc length can also be expressed in terms of the angle swept out by the arc: s = . Our expression for work in this simple case becomes: W = Frθ = τμ Since Fr gives us our torque, we can simplify our expression in terms of only τ and μ . What if the force is not perpendicular to the radius of the particle? Let the angle between the force vector and the radius vector be θ , as shown below. Figure %: A force acting at angle θ to the radius of rotation of point P
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Unformatted text preview: To compute the work we calculate the component of the force acting in the direction of the particle's displacement. In this case, this quantity is simply F sin . Again, this force acts over an arc length given by r . Thus the work is given by: W = ( F sin )( r ) = ( Fr sin ) Recall that = Fr sin Thus W = Surprisingly enough, this equation is exactly the same as our special case when the force acted perpendicular to the radius! In any case, the work done by a given force is equal to the torque it exerts multiplied by the angular displacement. For you calculus types, there is also an equation for work done by variable torques. Instead of deriving it, we can just state it, as it is quite similar to the equation in the linear case: W = d Thus we have quickly gone through deriving our expression for work. The next thing after work we studied in linear motion was kinetic energy, and it is to this topic that we turn....
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