An object is formed by attaching a uniform, thin rod with a mass of m r = 7.07 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.07 kg and length L = 5.2 m to a uniform sphere with mass ms = 35.35 kg and radius R = 1.3 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 = 426 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 = 426 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

(1)

moment of inertia of the rod along the given axis = m*L2/3 = [7.07*5.2*5.2/3] = 63.72 Kg.m2

moment of inertia of the sphere along the given axis = 2/5*m*r2 + m*x12 = [2/5*35.35*1.3*1.3 ] + [35.35*6.5*6.5]= 1517.42 Kg.m2

moment of inertia of the object = 63.72 +1517.42 = 1581.14 Kg.m2. = I(left)

(2)

Torque = force *distance

= 426*5.2/2 = 1107.6 N.m

angular acceleration * moment of inertia = torque

angular acceleration = 1107.6 / 1581.14 = 0.70 rad /s2

(3)

center of mass = [2.6*7.07 + 35.35*6.5] / 35.35+7.07   

= 5.85 m from the left end.

moment of inertia of rod = m*L2/12 + m*x22 = [7.07*5.2*5.2/12] + [7.07*3.25*3.25] = 90.60 kg.m2.

moment of inertia of sphere = 2/5*m*r2 + m*x32= [2/5*35.35*1.3*1.3 ] + [35.35*0.65*0.65] = 38.82 kg.m2.

moment of inertia about center of mass = 90.60 + 38.82 = 129.42 kg.m2. = I(CM)

(4)

if the force is applied parallel to the rod then cross product between the force vector and the distance vector is 0 degrees.

hence no torque formed

so no angular acceleration.

(5)

moment of inertia of rod = m*L2/12 + m*x42= [7.07*5.2*5.2/12] + [7.07*5.2*5.2] = 207.10 kg.m2

moment of inertia of sphere = 2/5*m*r*r + mx5*x5 = [2/5*35.35*1.3*1.3 ] + [35.35*1.3*1.3] = 83.63 kg.m2.

total moment of inertia = 207.10 + 83.63 = 290.73 Kg.m2 = I (right)

(6)

I (CM) < I (right) < I (left)

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