an object is formed by attaching a uniform thin rod with a mass of mr=6.61 kg an
ID: 1262088 • Letter: A
Question
an object is formed by attaching a uniform thin rod with a mass of mr=6.61 kg and length L=5.28 m to a uniform sphere with mass ms= 33.05 kg and radius R=1.32 m Note ms=5 mr 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 obect is fixed at the left end of the rod what is the angular acceleration if a force F=489 N is exerted perpendicular to the rod at the center of the rod?
3.what is the moment of inertia of the obect about an axis at the center of mass of the obect 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 obect is fixed at the center of mass what is the angular acceleration if a force F=489 N is exerted parallel to the rod at the end of the rod?
5. what is the moment of inertia of the object about an axis at the right edge of the sphere?
Explanation / Answer
(1):
MOI of the system = moi of rod at left end + moi of sphere at it's center + mass of sphere * (length of rod + radius of sphere )^2
MOI of the system = (1/3)*m*L^2 + (2/5)*5m*R^2 + 5m* (L+R)^2
put the values and get the answers.
(2):
angular acceleration = [489 * (L/2) ] / MOI of system
angular acceleration = [489 * (L/2) ] /{(1/3)*m*L^2 + (2/5)*5m*R^2 + 5m* (L+R)^2}
put the values and get the answer
(c):
center of mass = (m*L/2 +5m*L ) / ( 6m) = (5/6)L
moi accross center of mass = moi accross left end of rod + mass of system * ((5/6)L)^2
= ((1/3)*m*L^2 + (2/5)*5m*R^2 + 5m* (L+R)^2) + 6m*((5/6)L)^2
(d):
angular acceleration = 0
because perpendicual distance is zero
(e):
MOI accross the edge of the sphere = moi of the system accross the left end oof rod + mass of system * (L+2R)^2
= (1/3)*m*L^2 + (2/5)*5m*R^2 + 5m* (L+R)^2 + 6m * (L+2R)^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.