A thin steel plate is in the shape of a half circle

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SUMMARY

The discussion focuses on calculating various properties of a thin steel plate shaped as a half circle with a radius of 4 meters and a surface density of (3+r) kg/m². The calculations required include the area, mass, center of mass, and rotational inertia about specified axes. The area was correctly calculated as 8π square meters. Further assistance is requested for the mass, center of mass, and rotational inertia calculations.

PREREQUISITES
  • Understanding of calculus, particularly integration techniques.
  • Familiarity with concepts of mass distribution and density.
  • Knowledge of rotational inertia and its calculation methods.
  • Basic principles of mechanics, including motion equations.
NEXT STEPS
  • Calculate the mass of the half circle using the surface density function.
  • Determine the center of mass using the continuous mass distribution formula.
  • Compute the rotational inertia about the j-axis using the provided equations.
  • Analyze rotational inertia about an axis parallel to the j-axis through the center of mass.
USEFUL FOR

Students studying physics or engineering, particularly those focusing on mechanics and materials, as well as educators seeking to enhance their understanding of mass distribution and rotational dynamics.

Melina
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Homework Statement


The half thin steel plate has a radius of 4 meters and a surface density of (3+r) kg/m^2, where r is the radial distance from the origin. Using calculus, find:
A. its area
B. its mass
C. Its center of mass with respect to the origin shown,
D. It's rotational inertia about the j axis
E. Its rotational inertia about an axis parallel to the j axis and passing through the center of mass.

Homework Equations


d=v0 t + (1/2)at^2
a=v^2 / r
v=v0 + at
v=dx/dt
a= dv/dt
σ = dm/dA φ = dm/dV
I = r^2 dm Ix+Iy = Iz
I = ICM + m k^2 (CM:center of mass subscript)
Discrete masses:
xCM = ∑mixi (CM:center of mass subscript)
i
-----------
∑mi
i
continuous mass distributions:
xCM = 1/M ∫x dm (CM:center of mass subscript)

The Attempt at a Solution


A. A= 1/2 pi R^2 =8pi
B.
C.
D.
E.
 

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Part A is right. Please show your attempts for the other parts if you want help.
 

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