From inspection of the torque equation we can find another quantity called the moment of inertia,denoted I, which is the "angular mass".I = mr2 = moment of inertiaThis was for one point-like mass in the example, but what if we had two balls on the same stringsome distance apart from one another. Well like before each of these point-like masses has a torqueand we are allowed to sum all our torques (similar to forces), i.e.Tnet = T1+ 72 = (mıraı + m2ržaz2)2Then our moment of inertia for this system would beI = mır} + m2rWhat if we kept adding masses along the string? Well eventually we would have a distribution ofmasses along this string that would have to be summed for each radius squared distance they areaway from our pivot. If we made a line of masses then we would need to integrate over thisdistribution of infinitesimal masses to find the total moment of inertia of our system, i.e.Tneta fr² dm 2Let's not worry too much about how one would go about integrating out a distribution of masses, fornow it is sufficient to say that we need to add up each mass multiplied by its radius arm squaredfrom the pivot. The consequence of this is that you need more torque to move a heavier object (e.g.rolling a large car tire is more difficult than a bicycle tire). Lets come back to the case of ball attached at the end of the string. Suppose you have a long stringwith three balls attached at different lengths along the string. Ball 1 has a mass of 100 grams and isattached at a length of 10 cm. Ball 2 has a mass of 150 grams and is also attached at a length of 10cm. Ball 3 is 120 grams and is attached at a length of 8 cm. Assume the string's mass is negligible.You began whirling the 3 balls holding the end of the string. What is the total moment of inertia ofthe entire system?3.268 x 10kg. m23.268 x 10-4 kg. m2O 4.258 x 10-4 kg. m2O 4.258 x 10-3 kg. m2

Moment of Inertia The moment of inertia of a body about an axis is the sum of the product of the…

Answered: Lets come back to the case of ball…

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