An object is formed by attaching a uniform, thin rod with a mass of m r = 7.03 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.03 kg and length L = 5.12 m to a uniform sphere with mass ms = 35.15 kg and radius R = 1.28 m. Note ms = 5mr and L = 4R.

1)What is the moment of inertia of the object about an axis at the left end of the rod? kg-m2

2)If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 415 N is exerted perpendicular to the rod at the center of the rod? rad/s2

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.) kg-m2

4)If the object is fixed at the center of mass, what is the angular acceleration if a force F = 415 N is exerted parallel to the rod at the end of rod? rad/s2

5)

What is the moment of inertia of the object about an axis at the right edge of the sphere? kg-m2

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

moment of inertia of sphere about its center Icm = 2ms*r^2/5

using parallel theorem moment of inertia of sphere about end of rod

Is = ms(r+L)^2 + 2msr^2/5

I = Is + Irod = ms(r+L)^2 + 2msr^2/5 + mr*L^2/3

I = 1439.744 + 23.035904 + 61.429 = 1524.208981

2)

t = Fr = Ialpha

alpha = I/Fr

r = L/2

alpha = 1.435 rad/s^2

part 3 )

moment of inertia of rod about its center of mass = mr*L^2/12

parallel axis theorem for the rod gives = Ir = Ircm + mr(L/2 + r/2)^2

parallel axis theorem for the sphere gives = Is = Iscm + ms(r/2)^2

I = Ir + Is = mr*L^2/12 + mr(L/2+r/2)^2 + 13*msr^2/20

I = 15.3573 + 71.9872 + 37.433344 = 124.777844 kg.m^2

part 5 )

again use parallel axis theorem

Is' = Iscm + msr^2 = 7msr^2/5

Ir' = Ircm + mr(L/2 + 2r)^2 = mr(2r + L/2)^2 + mr*L^2/12

I = Is' + Ir'

I = mr(2r + L/2)^2 + 7ms*r^2/5 + mr*L^2/12

I = 184.287232 + 80.625664 + 15.35726933

I = 280.2701653 kg.m^2

6 ) ICM < Iright < Ileft

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