In the following problems, assume the following parameters: Mass of the bottom s
ID: 1425873 • Letter: I
Question
In the following problems, assume the following parameters:
Mass of the bottom steel disc: 1.4678 kg.
Mass of the upper steel disc: 1.4795 kg.
Mass of the upper aluminum disc: 0.4614 kg.
Radius of each disc: 63.3 mm.
Radius of the take-up spool: r=1.25 cm.
Mass of the falling weight: 25 grams.
1. Using the values given above, calculate the following:
a. The moment of interia of the lower, steel disc. (Answer: 2.941 * 10^-3 kg/m^2)
b. The moment of intertia of the upper, steel disc. (Answer: 2.964 * 10^-3 kg/m^2)
c. The moment of intertia of the upper, aluminum disc. (Answer: 9.244 *10^-4 kg/m^2)
d. The combined moment of inertia of the two steel discs. (Answer: 5.911 * 10^-3 kg/m^2)
2. Using the results of problem 1, what percentage error is made by neglecting mr^2 with respect to I in equation (100)?
Equation 100= mgr/(mr^2+I)
3. Suppose that the counter is reading 400 counts per second at some moment. What is the value of w at that moment?
NOTE: the register updates every two seconds.
I JUST NEED HELP WITH NUMBER 2 AND 3, I'M CONFUSED ON WHAT THE MASSES AND RADII SHOULD BE IN THE EQUATIONS
Explanation / Answer
(2) For the equation 100 = mgr/(mr^2+I)
Use m = 1.4678 + 1.4795 + 0.4614 = 3.4087 kg
(3) For the value of w at that moment,
Use w = mg
where m = 1.4678 + 1.4795 + 0.4614 + 0.025 = 3.4337 kg
=> w = mg = (3.4337(*(9.8) = 33.65026 N
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