93 Experiment 9. Moment of Inertia 9.5 Preliminary Assignment A Pasco rotational
ID: 1453794 • Letter: 9
Question
93 Experiment 9. Moment of Inertia 9.5 Preliminary Assignment A Pasco rotational motion apparatus with a step pulley diameter of (3.600 ± 0.003) cmis used to determine the experimental value for the moment of inertia of a ring. 1) The mass needed to overcome the friction of the axle with just the platter alone is (10.4±0.1) g . The hanging mass used to measure the acceleration of the platter is (70.4±0.1)g. With these masses the resulting velocity vs. time graph is fitted with a straight line, which yields the following fit result: V(t)-(0.0183-1 + 0.0036)% Calculate the moment of inertia of the rotational motion apparatus (without uncertainty) for the platter alone (pn)using equation (9.9). A. B. The mass needed to overcome the friction of the axle for the platter with the ring is 16.5±0.1 g The hanging mass used to measure the acceleration of the platterwith the ring is (70.4±0.1) g . With these masses the resulting velocity vs. time graph is fitted to a straight line, which yields the following fit result: V(t)-(0.0128-t + 0.0042)%. Calculate the moment of inertia ofthe rotational motion apparatus (without uncertainty) for the platter with the ring () using equation (9.9) C. Calculate the experimental value for the moment of inertia of the ring (without uncertainty) using equation (9.10). The inner diameter ofthe ring is (11.320 ±0.003)cm while the outer diameter is (13.598 ±0.003)cm. The mass of the ring is measured to be (731.2±0.1)s. Calculate the moment of inertia of the ring (with uncertainty) using equation (9.7). 2)Explanation / Answer
1) a) moment of inertia = 0.0704 * 0.0182 * ((9.8/0.0183) - 1)
= 0.01219 kg.m2
b) moment of inertia = 0.0704 * 0.0182 * ((9.8/0.0128) - 1)
= 0.01744 kg.m2
c) experimental value = 0.01744 - 0.01219 = 5.25 * 10-3 kg.m2
2) moment of inertia = 1/2 * 0.7312 * (0.05662 + 0.067992)
= 2.861 * 10-3 kg.m2
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