An object is formed by attaching a uniform, thin rod with a mass of m r = 6.98 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.98 kg and length L = 5.04 m to a uniform sphere with mass ms = 34.9 kg and radius R = 1.26 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 = 401 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 = 401 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

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.98 kg and length L = 5.04 m to a uniform sphere with mass ms = 34.9 kg and radius R = 1.26 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 = 401 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 right edge of the sphere? kg-m2 6) Compare the three moments of inertia calculated above: ICM

Explanation / Answer

Given data,

mr = 6.98 kg   L = 5.04 m
ms = 34.9 kg   R = 1.26 m
it implies ms = 5mr and L = 4R.

[1] Moment of inertia of the sphere and the rod system about an axis that is passing through the free end

I = mr *(L^2)/3 + ms*[(2/5)R^2+(R+L)^2] (By applying parallel axis theorem)

   I = 1466.45 kg m^2

[2] torque = I * angular accleration

   401 * 5.04/2 = 1466.45 * angular accleration

   angular accleration = 0.689 rad/sec^2

[3] From the center of the rod the COM of the sphere and rod system would be at

X= (3R*5m)/(m+5m) =2.5R

therefore I through an axis passing throug that point would be

I = mr [*(L^2)/3 +(0.5R)^2] + ms*[(2/5)R^2+(0.5R)^2] (By applying parallel axis theorem)

I= 100.656 kg m^2

[4] similar to 2 nd question

[5]   I = mr [*(L^2)/3 +(2R)^2] + ms*[(2/5)R^2+(R)^2]

I= 180.9969 Kg m^2

[6] therefore
     ICM < Iright < Ileft   

     

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