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3 Physics problems: Inertia of object, cable tension, force on a hinge, torque, direction

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1.)
Let thinner and thicker parts have masses m1 and m2 respectively and volumes v1 and v2 respectively.
v1 = (L/2)*pi*(2a)^2 = 2*pi*L*a^2
v2 = (L/2)*pi*a^2 = (1/4)v1
or,
v1:v2 = 1:4
therefore,
m1 = M*1/(1+4) = M/5
and,
m2 = 4M/5

i.)
Ix = (1/2)*(M/5)*a^2 + (1/2)*(4M/5)*(2a)^2
=> Ix = 17*M*a^2/10 --Answer

ii.)
Iy = I1c + m1*x1c^2 + I2c + m2*x2c^2 + Ix/2
where, I1c and I2c are the moment of inertias of thinner and thicker part about their centroid assuming them thin rods. x1c and x2c are the distances of centroid of each part from the y axis. Ix/2 is added because of perpendicular theorem which says
for 2-dimensional objects like thin disk:
Ix = Iy + Iz = 2*Iy (in case of symmetric circular ...

Solution Summary

The solution shows all the workings to arrive at the answers to the problems.

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