Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 10 - Exercises and Problems - Page 194: 79


$I =5.2\times10^{-5} \ kgm^2$

Work Step by Step

We know that there are two forces on the mass. Thus, we find: $T=mg-ma$ We also can solve for acceleration: $\Delta y = \frac{1}{2}at^2$ Since the change in height is one, this becomes: $a=\frac{2}{t^2}$ Finally, we find: $\tau = I\alpha=bT$ $ I\alpha=bT$ $\frac{Ia}{r}=b(mg-ma)$ $\frac{Ia}{b}=b(mg-ma)$ $\frac{2Ia}{b^2t^2}=(mg-ma)$ $\frac{2I}{b^2t^2}=(mg-m\frac{2}{t^2})$ Thus, to get a line of slope I, we can graph $\frac{2}{b^2t^2}$ versus $(mg-m\frac{2}{t^2})$, or we could graph $\frac{2}{t^2}$ versus $b^2(mg-m\frac{2}{t^2})$. Doing this, we verify the moments of inertia. (When doing this, do not forget to subtract $I_0$ from the value of the slope.) Finally, we find their moment of inertia together to be $I =5.2\times10^{-5} \ kgm^2$ by graphing the best fit trendline and by taking the slope to be the moment of inertia.
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