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2. An object with moment of inertia ????1 = 9.7 x 10−4 kg ∙ m2 rotates at a speed of 3.0 rev/s. A 20 g mass with moment of inertia ????2 = 1.32 x 10−6 kg ∙ m2 is dropped onto the rotating object at a distance of 5.0 cm from the center of mass. What is the angular velocity of the combined object and mass after the drop?

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MechEngineer

Answer:

2.85 rad/s

Explanation:

5 cm = 0.05 m

20 g = 0.02 kg

When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:

[tex]I_2 = 1.32\times10^{−6} + 0.02 * 0.05^2 = 0.513\times10^{-4} kgm^2[/tex]

So the total moment of inertia of the system of 2 objects after the drop is:

[tex]I = I_1 + I_2 = 9.7\times10^{-4} + 0.513\times10^{-4} = 0.0010213 kgm^2[/tex]

From here we can apply the law of angular momentum conservation to calculate the post angular speed

[tex]\omega_1 I_1 = \omega_2 I[/tex]

[tex]\omega_2 = \omega_1 \frac{I_1}{I} = 3 \frac{9.7\times10^{-4}}{0.0010213} = 2.85 rad/s[/tex]

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