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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
My PLUS Activity
Note-taking
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Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with
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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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What is the moment of inertia of a hoop of mass M and radius R rotated about a
cylinder
axis, as shown below?
A hoop of radius R
Fortunately, we do not need to use calculus to solve this problem. Notice that
all the mass is the same distance R from the axis of rotation. Thus we do not
need to integrate over a range, but can calculate the total moment of inertia.
Each small element dm has a rotational inertia of R2dm, where r is constant.
Summing over all elements, we see that I = R2dm = R2M. The sum of all the
small elements of mass is simply the total mass. This value for I of MR2
agrees with experiment, and is the accepted value for a hoop.
Problem :
What is the rotational inertia of a solid cylinder with length L and radius R,
rotated about its central axis, as shown below?
A cylinder being rotated about its axis
To solve this problem we split the cylinder into small hoops of mass dm, and
width dr:
A cylinder being rotated about its axis, shown with a small element of mass
from the cylinder
This small element of mass has a volume of (2Πr)(L)(dr), where dr is the
width of the hoop. Thus the mass of this element can be expressed in terms of
volume and density:
dm = ρV = ρ(2ΠrLdr)
We also know that the total volume of the entire cylinder is given by: V = AL = ΠR2L. In addition, our density is given by the total mass of the cylinder
divided by the total volume of the cylinder. Thus:
ρ = =
Substituting this into our equation for dm,
dm = = 2rdr
Now that we have dm in terms of r, we simply have to integrate over all possible
values of r to get our rotational inertia:
I
=
r2dm
=
2r3dr
=
[r4/2]0R
=
Thus the rotational inertia of a cylinder is simply . Once
again, it has the form of kMR2, where k is some constant less than one.