An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.62 k

Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.62 kg and length L = 5.32 m to a uniform sphere with mass m_s = 33.1 kg and radius R = 1.33 m. Note m_s = 5m_r and L = 4R. What is the moment of inertia of the object about an axis at the left end of the rod? If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 493 N is exerted perpendicular to the rod at the center of the rod? What is the moment of inertia of the object about an axis at the center of mass of the object? If the object is fixed at the center of mass, what is the angular acceleration if a force F = 493 N is exerted parallel to the rod at the end of rod? What is the moment of inertia of the object about an axis at the right edge of the sphere?

Explanation / Answer

given :
mass of the rod=mr=6.62 kg

length of the rod=L=5.32 m

mass of the sphere=ms=33.1 kg

radius=R=1.33 m

moment of inertia of the rod about an axis at its edge is given by mr*L^2/3=62.45 kg.m^2

moment of inertia of the sphere about an axis passing through its center=2*ms*R^2/5=23.42 kg.m^2

distance of this axis from the given axis as per the question (axis at the left end of the rod)=R+L=6.65 m

using parallel axis theorem,

moment of inertia of the sphere about the new axis=moment of inertia about old axis+mass*distance between axis^2

=23.42+33.1*6.65^2=1487.18 kg.m^2

so net moment of inertia=62.45+1487.18=1549.63 kg.m^2

part 2:

torque applied=force*distance from the axis

=493*(L/2)=1311.38 N.m

as we know, torque=angular acceleration*moment of inertia

==>1311.38=1549.63*angular acceleration

==>angular acceleration=0.8462 rad/sec^2

part 3:

center of mass is at a distance of R/2 from the right edge of the rod and R/2 from the center of the sphere

hence moment of inertia of the rod=62.45+mr*(R/2)^2=65.38 kg.m^2

moment of inertia of the sphere=23.42+ms*(R/2)^2=38.05 kg.m^2

hence total moment of inertia=65.38+38.05=103.43 kg.m^2

part 4:

as the force is parallel to the rod and passes through the center of mass,

torque=0

hence angular acceleration=0 rad/s^2

part 5:

right edge of the sphere is at a distance of 2*R from the right edge of the rod

and a distance of R from the center of the sphere.

hence moment of inertia of the rod=62.45+mr*(2*R)^2=109.29 kg.m^2

moment of inertia of the sphere=23.42+ms*(R)^2=81.97 kg.m^2

hence total moment of inertia=109.29+81.97=191.26 kg.m^2

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