An object is formed by attaching a uniform, thin rod with a mass of mr-7.29 kg a

Question

An object is formed by attaching a uniform, thin rod with a mass of mr-7.29 kg and length L-5.64 m to a uniform sphere with mass ms-36.45 kg and racius R-1.41 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? g-m2 Submit 2)lf the object is fixed at the left end of the rod, what is the angular accelerai if a force F- 494 N is exerted perpendicular to the rod at the center of the rod? Submit 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.) 4) lf the object is fixed at the center of mass, what is the angular acceleration if a force F- 494 N is exerted parallel to the rod at the end of rod? 2 Submit 5) What is the moment of inertia of the object about an axis at the right edge of the sphere? kg-m2 Submit

Explanation / Answer

1) Using parallel axis theorem,

I= Irod +Isphere,cg +ms x (rs+lrod)2

=7.29*5.64*5.64/3 + 0.4*36.45*1.41*1.41 +36.45*(5.64+1.41)2  kg-m.m

=77.297 + 28.986 + 1811.656 = 1917.94 kg-m.m

2) torque= I x alpha

=> alpha=angular acc = T/I=(494*5.64/2)/1917.94 = 0.7263 rad/s2

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