An object is formed by attaching a uniform, thin rod with a mass of m r = 7.43 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.43 kg and length L = 4.92 m to a uniform sphere with mass ms = 37.15 kg and radius R = 1.23 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 = 434 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 = 434 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

Plzzzzzzzz I need help with this question! :(

Explanation / Answer

mr = 7.43 L = 4.92
ms = 37.15 and R = 1.23 m.
ms = 5mr and L = 4R.

What is the moment of inertia of the object about an axis at the left end of the rod?
Where is the object attached, to the left or to the right of the rod?
Assuming that the sphere is attached to the right end of the rod

MI of the sphere about the left 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* 7.43*1.23^2 = 981.6 kg.m^2

What is the moment of inertia of the object about an axis at the left end of the rod?
= I
F*d = I
434*4.92/2 = 981.6
= 1.08 rad/s^2

Lets calculate moment of inertia of the object about an axis at the center of mass of the object

The c.m is at ( L+R)/2 from the left end = 5R/2.
Icm = Iz –md^2 = 981.6 - (5R/2.)^2 = 981.6 - (2.5*1.23)^2= 972.14 kg m^2

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