An object is formed by attaching a uniform, thin rod with a mass of m r = 7.21 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.21 kg and length L = 5.48 m to a uniform sphere with mass ms = 36.05 kg and radius R = 1.37 m. Note ms = 5mr and L = 4R.

3) What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

5) What is the moment of inertia of the object about an axis at the right edge of the sphere?

6) Compare the three moments of inertia calculated above:

ICM < Ileft < Iright

ICM < Iright < Ileft

Iright < ICM < Ileft

ICM < Ileft = Iright

Iright = ICM < Ileft

Explanation / Answer

mr = 7.21 L = 5.48
ms = 36.05 R = 1.37 m.
ms = 5mr and L = 4R.

5) moment of inertia of the object about an axis at the right edge of the sphere.

MI of the sphere about the right end of the rod,
= 2/5 ms*R^2 + ms* L^2
= 2mrR^2 +5mr*16R^2 = 82 mr.R^2

MI of the rod about the left end of the rod
= mrL^2/3 = 16mrR^2/3

MI of the system = 87.33 mr R^2 = 87.33* (7.21L)*(1.37)^2/3

=393.9L

=393.9*(4R)

=393.9*4*1.37

=2158.7kg.m^2

1)calculating the moment of inertia of the object about an axis at the center of mass of the object

as given the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

Distance is ( L+R)/2 from the left end = 5R/2.

Icm = Iz

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.