An object is formed by attaching a uniform, thin rod with a mass of m r = 7.26 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.26 kg and length L = 5.6 m to a uniform sphere with mass ms = 36.3 kg and radius R = 1.4 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 = 485 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 = 485 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?

PLEASE SHOW STEPS

Explanation / Answer

(1) We know that the moment of inertia of the sphere is (2/5)MR2 about its C.G.
Moment of inertia of rod about its end is ML2/3
So we have to take the moment of inertia about the one end of the rod
= mL2/3 + (2/5)MR2 + ML2
where m is the mass of rod , M is the mass of sphere,R is the radius od sphere and L is the length of the rod.
= (7.26)*5.62 /3 + (2/5)(36.3)(1.4)2 + (36.3)*(5.6)2
= 75.8912 + 1130.399 + 1138.368
= 2344.65 kg -m2
(2) We know that
torque = I*a where I is moment of inertia and a is angular acceleration
torque = Force*perpendicular distance
Torque = 485*(5.6/2) = 1358
put in the above equation
1358 = 2344.65*a
a = 0.5719 rad/s2

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