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Lets come back to the case of ball attached at the end of the string. Suppose you have a long string with three balls attached at different lengths along the string. Ball 1 has a mass of 100 grams and is attached at a length of 10 cm. Ball 2 has a mass of 150 grams and is also attached at a length of 10 cm. 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 of the entire system?

Lets come back to the case of ball attached at the end of the string. Suppose you have a long string with three balls attached at different lengths along the string. Ball 1 has a mass of 100 grams and is attached at a length of 10 cm. Ball 2 has a mass of 150 grams and is also attached at a length of 10 cm. 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 of the entire system?

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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 inertia This was for one point-like mass in the example, but what if we had two balls on the same string some distance apart from one another. Well like before each of these point-like masses has a torque and we are allowed to sum all our torques (similar to forces), i.e. Tnet = T1+ 72 = (mıraı + m2ržaz2)2 Then our moment of inertia for this system would be I = mır} + m2r What if we kept adding masses along the string? Well eventually we would have a distribution of masses along this string that would have to be summed for each radius squared distance they are away from our pivot. If we made a line of masses then we would need to integrate over this distribution of infinitesimal masses to find the total moment of inertia of our system, i.e. Tnet a fr² dm 2 Let's not worry too much about how one would go about integrating out a distribution of masses, for now it is sufficient to say that we need to add up each mass multiplied by its radius arm squared from 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).
Transcribed Image Text: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 inertia This was for one point-like mass in the example, but what if we had two balls on the same string some distance apart from one another. Well like before each of these point-like masses has a torque and we are allowed to sum all our torques (similar to forces), i.e. Tnet = T1+ 72 = (mıraı + m2ržaz2)2 Then our moment of inertia for this system would be I = mır} + m2r What if we kept adding masses along the string? Well eventually we would have a distribution of masses along this string that would have to be summed for each radius squared distance they are away from our pivot. If we made a line of masses then we would need to integrate over this distribution of infinitesimal masses to find the total moment of inertia of our system, i.e. Tnet a fr² dm 2 Let's not worry too much about how one would go about integrating out a distribution of masses, for now it is sufficient to say that we need to add up each mass multiplied by its radius arm squared from 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 string with three balls attached at different lengths along the string. Ball 1 has a mass of 100 grams and is attached at a length of 10 cm. Ball 2 has a mass of 150 grams and is also attached at a length of 10 cm. 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 of the entire system? 3.268 x 10 kg. m2 3.268 x 10-4 kg. m2 O 4.258 x 10-4 kg. m2 O 4.258 x 10-3 kg. m2
Transcribed Image Text:Lets come back to the case of ball attached at the end of the string. Suppose you have a long string with three balls attached at different lengths along the string. Ball 1 has a mass of 100 grams and is attached at a length of 10 cm. Ball 2 has a mass of 150 grams and is also attached at a length of 10 cm. 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 of the entire system? 3.268 x 10 kg. m2 3.268 x 10-4 kg. m2 O 4.258 x 10-4 kg. m2 O 4.258 x 10-3 kg. m2
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