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No Fear Translations of Shakespeare’s plays (along with audio!) and other classic works
Flashcards
Mastery Quizzes
Infographics
Graphic Novels
AP® Test Prep PLUS
AP® Practice & Lessons
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Testimonials from SparkNotes Customers
No Fear provides access to Shakespeare for students who normally couldn’t (or wouldn’t) read his plays. It’s also a very useful tool when trying to explain Shakespeare’s wordplay!
Erika M.
I tutor high school students in a variety of subjects. Having access to the literature translations helps me to stay informed about the various assignments. Your summaries and translations are invaluable.
Kathy B.
Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with understanding the crux of the text.
Kay H.
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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.
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 = rμ. 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
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 sinθ. Again, this force acts over an arc length given by rμ.
Thus the work is given by:
W = (F sinθ)(rμ) = (Fr sinθ)μ
Recall that
τ = Fr sinθ
Thus W = τμ
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 = τdμ
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.
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.