An object is formed by attaching a uniform, thin rod with a mass of m r = 7 kg a

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7 kg and length L = 5.28 m to a uniform sphere with mass ms = 35 kg and radius R = 1.32 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

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2)

If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 460 N is exerted perpendicular to the rod at the center of the rod?
rad/s2

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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

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4)

If the object is fixed at the center of mass, what is the angular acceleration if a force F = 460 N is exerted parallel to the rod at the end of rod?
rad/s2

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5)

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

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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)

I = (1/3)*mr*L^2 + (2/5)*ms*R^2 + ms*(L+R)^2

I = ((1/3)*7*5.28^2) + ((2/5)*35*1.32^2)+(35*(5.28+1.32)^2)

I = 1614.04 kgm^2

(2)

torque = I*alfa

F*L/2 = I*alfa

460*5.28/2 = 1416.04*alfa

alfa = 0.86 rad/s^2

(3)

I = (1/12)*mr*L^2 + mr*(L/2+R/2)^2 + (2/5)*ms*R^2 + ms*(R/2)^2

I = ((1/12)*7*5.28^2)+(7*(2.64+0.66)^2) + ((2/5)*35*1.32^2)+(35*0.66^2)

I = 132.13 kg m^2

(4)

alfa = F*(L/2+R/2)/I

alfa = (460*(2.64+0.66))/132.13 = 11.5 rad/s^2

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