Answer :
Final Answer:
The moment of inertia for the system, which includes a 28 g rod with two slugs at masses 81 g and 76 g located 16.7 cm from the pivot point, is approximately 0.00666 kg·m².
Explanation:
To find the moment of inertia for the given system, we'll need to consider the moment of inertia of each individual component (the rod, the 81 g slug, and the 76 g slug) and then sum them up to get the total moment of inertia.
First, let's calculate the moment of inertia for the rod. The formula for the moment of inertia of a slender rod rotating about an axis perpendicular to the length of the rod and passing through its center is:
I_rod = (1/12) * m_rod * L^2
Where:
I_rod = Moment of inertia of the rod
m_rod = Mass of the rod = 28 g = 0.028 kg (converted)
L = Length of the rod = 38.1 cm = 0.381 m (converted)
Now, we'll calculate the moment of inertia for the two slugs. The formula for the moment of inertia of a point mass rotating about an axis a distance r from the axis of rotation is:
I_point = m_point * r^2
For the 81 g slug:
m_1 = 81 g = 0.081 kg (converted)
r_1 = 16.7 cm = 0.167 m (converted)
For the 76 g slug:
m_2 = 76 g = 0.076 kg (converted)
r_2 = 16.7 cm = 0.167 m (converted)
Now, we can calculate the moment of inertia for each component:
I_rod = (1/12) * 0.028 kg * (0.381 m)^2
I_1 = 0.081 kg * (0.167 m)^2
I_2 = 0.076 kg * (0.167 m)^2
Finally, we sum up the individual moments of inertia to find the total moment of inertia for the system:
Total I = I_rod + I_1 + I_2
Total I ≈ 0.00666 kg·m²
So, the moment of inertia for the given system is approximately 0.00666 kg·m². This accounts for the rod and the two slugs.
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