I(end) = I(center of mass) + M(total)D^2
Solve for I(CM) = I(end) - M(total)D^2
M(total) = m(sphere) + m(rod)
D = Distance from end of rod to CM = L + R/2
I(end) = I(rod) + I(sphere)
I(rod) = 1/3(mass of rod)(Length of rod)^2
I(end, sphere) = 2/5(mass of sphere)(Radius)^2 + (mass of sphere)D^2
Here, D = distance from CM to left axis = L + R
An object is formed by attaching a uniform, thin rod with a mass of mr = 6.54 kg and length L = 5.12 m to a uniform sphere with mass ms = 32.7 kg and radius R = 1.28 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?
1417.969664
2)If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 467 N is exerted perpendicular to the rod at the center of the rod?
.8431209992
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 = 467 N is exerted parallel to the rod at the end of rod?
0
5)What is the moment of inertia of the object about an axis at the right edge of the sphere?
260.734976
All above answers are correct, Ive been having troubles with problem 3 for several hours I also talk to a bunch of TAs and even they couldn't help me out. If any one could help me out I would appreciate it. Thanks.
1 answer
2 answers