Find the inertia tensor if the origin is moved to the center of mass of the T-sq
ID: 2252436 • Letter: F
Question
Find the inertia tensor if the origin is moved to the center of mass of the T-square.
Suppose the T-square is allowed to pivot back and forth about the original origin. What is the angular frequency of the (small) oscillations?
A T-square is formed by joining two identical thin rods of length L = 4.6 m and mass m/2 (so that the total mass of the T-square is m = 22.4 kg.) Find the inertia tensor of this arrangement if the origin is placed at the point of contact between the two rods, as shown in the figure. Enter your 6 independent elements in this order: I11, I12, I13, I22, I23, I33. Find the inertia tensor if the origin is moved to the center of mass of the T-square. Suppose the T-square is allowed to pivot back and forth about the original origin. What is the angular frequency of the (small) oscillations?Explanation / Answer
Let subscript 1,2,3 corresponds to x,y,z
m = mass of 1 rod = 22.4 /2 = 11.2 kg ; Length = 4.6 m
Since, Rod is thin, thickness is ignored.
a) Ixx = MOI about x axis = m*L*L/3 + 0 = 79 kg.m2 [0 for the horizontal rod]
b) Iyy = MOI about y axis = m*L*L/12 = 19.75 kg.m2 [0 for vertical rod]
c) Izz = MOI about z axis = Ixx + Iyy = 98.75 kg.m2
d) I13 = Ixz = MOI about an axis in x-z plane and equally inclined to both the axis.
Vertical rod is along y axis and hence will have no change in MOI for any axis in x-z plane and for horizontal rod, the effective length of rod will be projection of rod perpendicular to axis.
Ixz = m*L*L/3 + m*Lcos 45*Lcos 45/12 = 88.872 kg.m2
3) Iyz
Same as before, No change in horizontal rod and projected length of vertical one.
Iyz = m*Lcos 45*Lcos 45/3 + m*L*L/12 = 59.248 kg.m2
4) I xy ; Projected length of both.
Ixy = m*Lcos 45*Lcos 45/12 + m*Lcos 45*Lcos 45/3 = 49.37 kg.m2
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