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Work - see that each one does in fact have a translational...

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Work, Energy and Combined Motion Having established the dynamics of rotational motion, we can now extend our study to work and energy. Given what we already know, the equations governing energetics are quite easy to derive. Finally, with the equations that we have derived, we will be able to describe the complicated situations involving combined rotational and translational motion. Rotational Kinetic Energy Consider a wheel spinning in place. Clearly the wheel is moving, and has a kinetic energy attached to it. But the wheel is not engaged in translational motion. How do we calculate the kinetic energy of the wheel? Our answer is similar to how we calculated the result of a net torque on a body: by summing over each particle. Given a rotating body, we state that the body is made up of n single rotating particles, each at a different radius from the axis of rotation. When each particle is considered individually, we can
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Unformatted text preview: see that each one does in fact have a translational kinetic energy: K = m 1 v 1 2 + m 2 v 2 2 + ... m n v n 2 However, we also know from our relation between linear and angular variables that v = rσ . Substituting this expression in, we see that: K = m 1 r 1 2 σ 2 + m 2 r 2 2 σ 2 + ... m n r n 2 σ 2 Since all particles are part of the same rigid body, we can factor our σ 2 : K = ( mr 2 ) σ 2 This sum, however, is simply our expression for a moment of inertia. Thus: K = Iσ 2 As we might expect, this equation is of the same form as our equation for linear kinetic energy, but with I substituted for m , and σ substituted for v . We now have rotational analogues for nearly all of our translational concepts. The last rotational equation that we need to define is power....
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