Q NUMBER 5 NEED TO BE ANSRWED An object is formed by attaching a uniform, thin r
ID: 1611314 • Letter: Q
Question
Q NUMBER 5 NEED TO BE ANSRWED
An object is formed by attaching a uniform, thin rod with a mass of mr = 7.32 kg and length L = 5.72 m to a uniform sphere with mass ms = 36.6 kg and radius R = 1.43 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?
1980.85 kg-m2
2) If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 401 N is exerted perpendicular to the rod at the center of the rod?
0.5789 rad/s2
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.)
162.16 kg-m2
4)If the object is fixed at the center of mass, what is the angular acceleration if a force F = 401 N is exerted parallel to the rod at the end of rod?
0 rad/s2
5)
Q What is the moment of inertia of the object about an axis at the right edge of the sphere?
kg-m2
Explanation / Answer
moment of inertia of the rod about an axis passing through its center of mass=(mr*L^2)/12
distance between the axis shown to the right of the sphere and the axis passing through the center of mass of the rod
=2*R+(L/2)
=2*R+(4*R/2)
=4*R
using parallel axis theorem,
moment of inertia of the rod about the axis shown in the image
=moment of inertia about axis through center of mass + mass*distance between two axes^2
=(mr*L^2/12)+mr*(4*R)^2
=(7.32*5.72^2/12)+7.32*(4*1.43)^2
=259.457 kg.m^2
moment of inertia of the sphere about an axis passing through its center of mass=0.4*ms*R^2
distance between two axes=R
so total moment of inertia of the sphere using parallel axis theorem
=0.4*ms*R^2+ms*R^2
=104.78 kg.m^2
so total moment of inertia of the system about the axis shown
=259.457+104.78
=364.2376 kg.m^2
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