An object is formed by attaching a uniform, thin rod with a mass of m r = 6.83 k
ID: 1551124 • Letter: A
Question
An object is formed by attaching a uniform, thin rod with a mass of mr = 6.83 kg and length L = 5.72 m to a uniform sphere with mass ms = 34.15 kg and radius R = 1.43 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 = 455 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 = 455 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
Explanation / Answer
(1)We knw that the moment of inertia , I = mL2/3 + (2/5)MR2 + M*(L + R)2
I = 6.83kg * (5.72m)2 / 3 + 34.15kg * [(1.43m)2 + (5.72m + 1.43m)2 = 1890 kg·m2
(2) Torque = I*alpha
455N * 2.86m = 1890kg·m2 * alpha
So, angular acceleration,alpha == 0.688 rad/s2
(3) Measured from the left end of the rod, the Center of Mass is at
x = (6.83kg * 2.86m + 34.15kg * (5.72m * 1.43m)) / (6.83 + 34.15)kg
x = 7.293 m
Again invoking the parallel axis theorem,
I = mL2/12 + m(L/2 + R/2)2+ (2/5)MR2 + M(R/2)2
I = 6.83kg*[(5.72m)2/12 + (2.86m + 0.7m)2] + 34.15kg*[(2/5)(1.43m)2 + (0.7m)2]
I = 105.2 kg·m2 + 34.66kg·m2 = 139.8 kg·m2
(4) There is no torque in this case, r x F = 0 (they're parallel).
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