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rmunoz
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Rotational Inertia, Need Help With Integrals!
In the figure below, a small disk of radius r = 2.00 cm has been glued to the edge of a larger disk of radius R = 4.00 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.50 103 kg/m3 and a uniform thickness of 4.60 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O?
http://www.webassign.net/halliday8e/art/images/halliday8019c10/image_n/nfg051.gif
I=\frac{MR\stackrel{2}{}}{2}
I=\intr\stackrel{2}{}dm
\rho= \frac{m}{v}
com= \frac{m1x1 + m2x2 + m3x3...m(n)x(n)}{M}
I=Icom + Mh\stackrel{2}{}
My initial attempt basically included the following steps:
m(circle1)= \pir1\stackrel{2}{} * width [.0046m] * density [1.5*10\stackrel{3}{}kg/m\stackrel{3}{}]
m(circle2)= \pir2\stackrel{2}{} * width [.0046m] * density [1.5*10\stackrel{3}{}kg/m\stackrel{3}{}]
=> I(com)= \frac{1}{2}MR\stackrel{2}{} Where R is the radius' of both circles added together
This clearly was the wrong approach and I'm fairly certain that in order to get the correct answer for this, i do not know how to take the integral (im only somewhat familiar with the process).
That integral is supposed to give me the I (com) and then I am fairly certain all i have to do is find Mh^2 and add the two together to get the I (sys).
Would anybody mind helping me with this process, by explaining how to take a simple integral (preferably using these exact same terms), and what exactly this is doing for the calculations. This kind of assistance would be much appreciated!
Homework Statement
In the figure below, a small disk of radius r = 2.00 cm has been glued to the edge of a larger disk of radius R = 4.00 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.50 103 kg/m3 and a uniform thickness of 4.60 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O?
http://www.webassign.net/halliday8e/art/images/halliday8019c10/image_n/nfg051.gif
Homework Equations
I=\frac{MR\stackrel{2}{}}{2}
I=\intr\stackrel{2}{}dm
\rho= \frac{m}{v}
com= \frac{m1x1 + m2x2 + m3x3...m(n)x(n)}{M}
I=Icom + Mh\stackrel{2}{}
The Attempt at a Solution
My initial attempt basically included the following steps:
m(circle1)= \pir1\stackrel{2}{} * width [.0046m] * density [1.5*10\stackrel{3}{}kg/m\stackrel{3}{}]
m(circle2)= \pir2\stackrel{2}{} * width [.0046m] * density [1.5*10\stackrel{3}{}kg/m\stackrel{3}{}]
=> I(com)= \frac{1}{2}MR\stackrel{2}{} Where R is the radius' of both circles added together
This clearly was the wrong approach and I'm fairly certain that in order to get the correct answer for this, i do not know how to take the integral (im only somewhat familiar with the process).
That integral is supposed to give me the I (com) and then I am fairly certain all i have to do is find Mh^2 and add the two together to get the I (sys).
Would anybody mind helping me with this process, by explaining how to take a simple integral (preferably using these exact same terms), and what exactly this is doing for the calculations. This kind of assistance would be much appreciated!
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