An object is formed by attaching a uniform, thin rod with a mass of mr = 6.8 kg

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.8 kg and length L = 5.64 m to a uniform sphere with mass ms = 34 kg and radius R = 1.41 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?

2) If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 445 N is exerted perpendicular to the rod at the center of the rod?

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.)

4) If the object is fixed at the center of mass, what is the angular acceleration if a force F = 445 N is exerted parallel to the rod at the end of rod?

5)What is the moment of inertia of the object about an axis at the right edge of the sphere?

Explanation / Answer

moment of inertia = moment of inertia of rod + moment of inertia of sphere

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

ms = 5mr

L = 4R

I = (1/3)*mr*16R^2 + (2/5)*5*mr*R^2 + 5*mr*(5R)^2

I = mr*R^2*[ (16/3) + (2) + 125 ]

I = 6.8*1.41^2*(((16/3) + (2) + 125 )

I = 1789.02 kg m^2

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2)

torque = (L/2)*F

but torque = I*alpha

(5.64/2)*445 = 1789.02*alpha

alpha = 0.701 rad/s^2 <<------answer

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3)

axis is at center of mass

from parallel axis theorem

I = Icm + M*r^2

M = mass of the object = ms + mr = 6mr = 5*6.8 = 40.8

r = perpendicular distance between parallel axis = (L+R/2)

r = (5.64+1.41/2) = 6.345 m

Icm = I - M*r^2

Icm = 1789.02 - (40.8*6.345^2)

Icm = 146.45 kg m^2 <<----answer

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torque = r*F*sintheta

here theta = 0

but torque = I*alpha

0 = 1789.02*alpha

alpha = 0 rad/s^2 <<------answer

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5)

I1 = Icm + M*r1^2

r1 = R/2 + R = 3R/2 = (3*1.41/2) = 2.115 m

I1 = 146.45 + (40.8*2.115^2)

I1 = 329 kg m^2

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