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GoldShadow
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Problem reads:
"A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes:
(a) an axis perpendicular to the bar through its center; (b) an axis perpendicular to the bard through one of its balls; (c) an axis parallel to the bar through both balls; (d) an axis parallel to the bar and 0.500m from it.
[tex]I=\sum m_{i}r^{2}_{i}[/tex]
[tex]I_{P}=I_{CM}+Md^{2}[/tex]
Moment of inertia for a slender rod, axis through center: [tex]I=\frac{1}{12}ML^{2}[/tex]
Moment of inertia for a slender rod, axis through one end: [tex]I=\frac{1}{3}ML^{2}[/tex]a) To find total moment of inertia, I used the equation [tex]I=\frac{1}{12}ML^{2}+2m(\frac{L}{2})^{2}[/tex], where "m" is the mass of one ball, "M" is the mass of the rod and "L" is the total length of the rod: [tex]I=\frac{1}{12}(4.00kg)(2.00m)^{2}+2(.500kg)(\frac{2.00m}{2})^{2}=2.33 kgm^{2}[/tex]
b) For this part I used [tex]I=\frac{1}{3}ML^{2}+mL^{2}[/tex], since one ball is on the axis of rotation and does not contribute to the moment of inertia. The answer I got:
[tex]I=\frac{1}{3}(4.00kg)(2.00m)^{2}+(.500kg)(2.00m)^{2}=7.33 kgm^{2}[/tex]
c) Zero, since the rod/balls are not given a radius; their distance from the axis of rotation is zero.
d) Using the parallel-axis equation, [tex]I_{P}=I_{CM}+Md^{2}[/tex], I had 0 for [tex]I_{CM}[/tex] as found in part C, 5.00 kg for "M" (total mass of rod and balls... or do I add each separately?) and .500m for "d" as given in the problem. My equation/answer:
[tex]I_{P}=(0)+(5.00kg)(0.500m)^{2}=1.25 kgm^{2}[/tex]
The answers in the back of my textbook are not only incomplete, they seem incorrect and I would appreciate any help. Thanks.
EDIT: I don't think I'm actually supposed to use the "parallel axis" equation in this problem on second look, so I'm positive I did part D wrong... but I'm still not entirely sure about parts a-c.
"A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes:
(a) an axis perpendicular to the bar through its center; (b) an axis perpendicular to the bard through one of its balls; (c) an axis parallel to the bar through both balls; (d) an axis parallel to the bar and 0.500m from it.
Homework Equations
[tex]I=\sum m_{i}r^{2}_{i}[/tex]
[tex]I_{P}=I_{CM}+Md^{2}[/tex]
Moment of inertia for a slender rod, axis through center: [tex]I=\frac{1}{12}ML^{2}[/tex]
Moment of inertia for a slender rod, axis through one end: [tex]I=\frac{1}{3}ML^{2}[/tex]a) To find total moment of inertia, I used the equation [tex]I=\frac{1}{12}ML^{2}+2m(\frac{L}{2})^{2}[/tex], where "m" is the mass of one ball, "M" is the mass of the rod and "L" is the total length of the rod: [tex]I=\frac{1}{12}(4.00kg)(2.00m)^{2}+2(.500kg)(\frac{2.00m}{2})^{2}=2.33 kgm^{2}[/tex]
b) For this part I used [tex]I=\frac{1}{3}ML^{2}+mL^{2}[/tex], since one ball is on the axis of rotation and does not contribute to the moment of inertia. The answer I got:
[tex]I=\frac{1}{3}(4.00kg)(2.00m)^{2}+(.500kg)(2.00m)^{2}=7.33 kgm^{2}[/tex]
c) Zero, since the rod/balls are not given a radius; their distance from the axis of rotation is zero.
d) Using the parallel-axis equation, [tex]I_{P}=I_{CM}+Md^{2}[/tex], I had 0 for [tex]I_{CM}[/tex] as found in part C, 5.00 kg for "M" (total mass of rod and balls... or do I add each separately?) and .500m for "d" as given in the problem. My equation/answer:
[tex]I_{P}=(0)+(5.00kg)(0.500m)^{2}=1.25 kgm^{2}[/tex]
The answers in the back of my textbook are not only incomplete, they seem incorrect and I would appreciate any help. Thanks.
EDIT: I don't think I'm actually supposed to use the "parallel axis" equation in this problem on second look, so I'm positive I did part D wrong... but I'm still not entirely sure about parts a-c.
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