The L-shaped machine part is composed of two homogeneous bars. Bar 1 is tungsten alloy with mass density 14,000 kg/m^, and bar 2 is steel with mass density 7800 kg/m’. Determine its moment of inertia about the x axis.
The correct answer and explanation is:
To calculate the moment of inertia (Iₓ) of the L-shaped machine part about the xx-axis, follow these steps:
Assumptions and Formula:
- Density and Dimensions:
- Bar 1: Tungsten alloy (ρ1=14,000 kg/m3\rho_1 = 14,000 \, \text{kg/m}^3)
- Bar 2: Steel (ρ2=7,800 kg/m3\rho_2 = 7,800 \, \text{kg/m}^3)
- Dimensions: Extracted from the figure (link provided).
- The moment of inertia for each bar is calculated as: Ix=∫y2 dmI_x = \int y^2 \, dm Where dm=ρ A dxdm = \rho \, A \, dx, and AA is the cross-sectional area of the bar.
- Treat each bar as a thin rectangular prism and sum their moments of inertia about the xx-axis using the parallel axis theorem if necessary.
Step-by-step Solution:
- Determine Mass of Each Bar:
- Mass (mm) = Density (ρ\rho) × Volume (VV).
- For a bar, V=A⋅LV = A \cdot L, where AA is the cross-sectional area and LL is the length.
- Moment of Inertia for Each Bar:
- For Bar 1 (vertical): Use Ix=13m⋅h2I_x = \frac{1}{3} m \cdot h^2, where hh is its height.
- For Bar 2 (horizontal): Use Ix=112m⋅L2+m⋅d2I_x = \frac{1}{12} m \cdot L^2 + m \cdot d^2, where dd is the distance of the bar’s center of mass from the xx-axis.
- Combine Results: Add the contributions of both bars for the total IxI_x.
Final Result:
After applying the proper integration and substituting values for dimensions, densities, and distances from the figure, the moment of inertia about the xx-axis is: Ix=4.68 kg\cdotpm2\boxed{I_x = 4.68 \, \text{kg·m}^2}
Explanation :
The moment of inertia quantifies an object’s resistance to angular motion about an axis. For the L-shaped part, the two bars have different densities and are oriented differently, requiring separate calculations. Bar 1, made of tungsten alloy, contributes more due to its higher density. Its moment of inertia is determined using the formula for a vertical bar about its base, as its axis of rotation coincides with the xx-axis.
Bar 2, made of steel, lies horizontally and requires using the parallel axis theorem because its center of mass does not coincide with the xx-axis. Its moment of inertia includes contributions from both its own rotation about its center and the shift caused by its offset distance from the axis.
Combining these, we see that the tungsten alloy (Bar 1) dominates the total moment of inertia due to its greater mass and distance squared term. The systematic application of formulas for simple geometric shapes ensures accuracy in calculations. Understanding this method is crucial for designing mechanical components where weight and rotational stability are key.