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A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considered. the first axis passes through the 50-cm mark and the second axis passes through the 30-cm mark. what is the ratio of the moment of inertia through the second axis to the moment of inertia through the first axis?

Respuesta :

Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m =  the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
[tex]I_{g} = \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} = \frac{m}{12} \, kg-m^{2} [/tex]

According to the parallel axis theorem, the moment of inertia about the first axis is
[tex]I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} = \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}[/tex]

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
[tex] \frac{I_{g}}{I_{1}} = \frac{0.0833m}{0.1233m} =0.6756[/tex]

Answer:  0.676

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JayDilla

Answer:  L2/L1=1.5

Explanation:

Not my work but heres the way the answer is obtained cause the other guy was totally incorrect

Write the expression to calculate the moment of inertia of the meter stick about second axis of rotation. I 2 = M L 2 12 + M ( x 1 − x 2 ) 2

Substitute the values in the above expression. I 2 = M ( 1 m ) 2 12 + M ( 0.5 m − 0.3 m ) 2 I 2 = M 12 + 0.04 M I 2 = 1.48 M 12

. . . . . . ( i i ) Divide expression (ii) by expression (i):

I 2 I 1 = 1.48 M 12 M 12 = 1.48 ≃ 1.5

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