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Mth 277
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Ch 12 Practice Test Question 6
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6. Show that a homogenous lamina of mass m that covers the circular region x^2 + y^2 = a^2 will have moment of inertia ma^2 / 4 with respect to both the x- and y-axes.
The region -1 <= x <= 1, -1 <= y <= 1 describes a square, not a circle.
What is the definition of moment of inertia and what needs to be integrated to find a moment
of inertia?
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I did a little further research and found the moments of inertia are:
Ix = double integral with function y^2 dydx
Iy = double integral with function x^2 dydx
The part that I am still confused on is the moment of inertia given the way it is. I know newton's second law is the torque = I*a and Force= mass*a so this could possibly be where the m and a come from but what I do not understand is what we are trying to do in this problem. I am assuming we are trying to set up a double integral again with dydx. However, do we use the x^2 + y^2 as the function and then would the bounds just be the -1 to 1 for both x and y? If you could help me set up this problem I would really appreciate it.
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You have the right expressions for I_x and I_y.
The concept is this:
The further a given amount of mass is from the axis of rotation the more torque it takes to provide a given amount of angular acceleration. This is because at a given angular velocity the mass will be moving faster if it is further from the axis, and will therefore gain more kinetic energy. The speed of the mass at a given angular velocity in fact being directly proportional to the distance from the axis, and the kinetic energy being proportional to the square of this distance, it turns out that the energy required to achieve a given angular velocity is proportional to the square of that distance. To achieve a given angular velocity (starting from rest) within a given time interval therefore requires a torque that varies with the square of this distance. Since the torque to achieve a given angular acceleration is proportional to what we wish to call the moment of inertia, the moment of inertia of the mass is proportional to the square of its distance from the axis of rotation.
Note that we use alpha, not a, for angular acceleration so the appropriate statements of Newton's Second Law are F = m a and torque = I * alpha.
This is a very abbreviated explanation of the physics of the situation. A full explanation would go into the distinction between linear and angular acceleration (which are related by the definition of the radian). If you want a more extensive explanation, you're welcome to ask.
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more to the point of this course:
You are asking the right questions and making some progress, but you don't yet appear to have developed the concept of describing a region with inequalities, and how this relates to setting up an integral. This is essential for nearly any application of integration, and it's necessary to almost everything that follows in this chapter and the next. I do suggest that you at least read through the qa's for sections 1 and 2, where the process is covered in detail.
If x goes from -1 to 1, then y does not. For a given x, y goes from the corresponding lower limit of the circular region to the other. If x happens to be zero, y would go from -1 to 1 while staying in the circle. But if x has any other value, the y values -1 and 1 would result in points (x, y) which are not within the circle.
Of course you could let y go from -1 to 1, in which case the limits on x would depend on y.
There are a number of examples in the text and in the posted q_a_'s (for sections 1 and 2) that address the limits on a region.
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