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A thin uniform bar has two small balls glued to its ends. The bar is 2.00 $\mathrm{m}$ long and has mass $4.00 \mathrm{kg},$ while the balls each have mass 0.500 $\mathrm{kg}$ 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 bar through one ofthe balls; (c) an axis parallel to the bar through both balls.

$| 2.33 \mathrm{kg} . \mathrm{m}^{2}$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Sheffield

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in the first part of this question, we're going to get a lead the moment off in that show off the system, about an excess perpendicular to the bar and the passing through the center sold at this moment off. In that shays, I want more. The moment off inertia of the system will be the summer home of Dolphin ASHA off the bar and the moment off in a show off the bus so we can write. I want people to I bought Plus I bought. So call it equating one. No using barrel exterior we can, right? I want people's to one by 12 ml squared. Bless him into Elba to all square. Hear Capital, Amis de Miser Bar and L is the Linda Block with the small Mr Preserved the Missiles balls while sitting the well use into this question we will get I want equals to one bite wealth into 4.0 crazy into 2.0 meter old square plus pull into a 0.500 K G into 1.0 meter old square. So this will give us I want people's too 2.33 kg meter square. So this is the answer off. First part of this question, let's come toward the second part off this question. In part B of this question, we are again going to kill. Let the moment off inertia off descent system to a nexus perpendicular to the bar. I am the passing the buck, passing the excess through one off the balls. So know this moment off in ashes, I too, using the barrel. Next year, we can right I to equal store. One bite left and we'll square blessed him into El by two whole square plus ml square so it can be simplify it is are two equals to one by three ml square plus camel square. So call it equation too. Now setting the Williams into this equation too. We will get I to equals to one by three into 4.0 k g into 2.0 meter hole square plus into 0.500 Casey in tow. 2.0 meter all square. So this will give us I ll equals two some 0.33 Did you meet? Oh, Square. Let's come toward the last part off this question in this part of the question. We're going to kill Claire. The moment off. Inertia off the same system about an exit. Spell it to the bar and a passing through the boat balls. Let this moment off. Initialize I three as we can see the moment off for this mass distribution is he goes to zero, that is articles to zero. So the moment off inertia should be close to zero. So three equals two zero k e meter square, so end off the question.

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