An object is formed by attaching a uniform, thin rod with a mass of m-7.31 kg an
ID: 2038622 • Letter: A
Question
An object is formed by attaching a uniform, thin rod with a mass of m-7.31 kg and length L 5.68 m to a sphere with mass m, -36.55 kg and radius R - 1.42 m. Note m,- 5m, and L- 4R. 1) What is the moment of inertia of the object about an axis at the left end of the rod? 1947.92 2) If the object is fixed at the left end of the rod, what is the anguiar acceleration if a force F-400 N is exerted perpendicular to the rod at the center of the rod? .725 3) rad/s Submit What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of can be calculated to be located at a point haifway between the center of 26.82 the sphere and the left edge of the sphere.) 4) If the object is flxed at the center of mass, what is the angular acceieration if a force F-400 N s exerted paralel to the rod at the end of rod? rad/s Submit 5) What is the moment of inertia of the object about an axis at the right edge of the sphere 6) Compare the three moments of inertia calculated above:Explanation / Answer
1)
I_left = mr*L^2/3 + (2/5)*ms*R^2 + ms*(L + R)^2
= 7.31*5.68^2/3 + (2/5)*36.55*1.42^2 + 36.55*(5.68 + 1.42)^2
= 1951 kg.m^2
2) Torque due to the force, T = r*F
= (5.68/2)*400
= 1136 N.m
angular acceleration, alfa = T/I
= 1136/1951
= 0.582 rad/s^2
3) let x is the distance between center of mass to center of the sphere.
ms*x = mr*(L/2 + R-x)
36.55*x = 7.31*(5.68/2 + 1.42 - x)
==> x = 0.71 m
Icm = (2/5)*ms*R^2 + ms*x^2 + mr*L^2/12 + ms*(L/2 + R - x)^2
= (2/5)*36.55*1.42^2 + 36.55*0.71^2 + 7.31*5.68^2/12 + 7.31*(5.68/2 + 1.42 - 0.71)^2
= 159.7 kg.m^2
4) T = r*F
= (L/2 + R - x)*F
= (5.68/2 + 1.42 - 0.71)*400
= 1420 N
angular acceleration, alfa = T/I
= 1420/159.7
= 8.89 rad/s^2
5) I_right = mr*L^2/12 + mr*(L/2 + 2*R)^2 + (2/5)*ms*R^2 + ms*(2*R)^2
= 7.31*5.68^2/12 + 7.31*(5.68/2 + 2*1.42)^2 + (2/5)*36.55*1.42^2 + 36.55*(2*1.42)^2
= 579.8 kg.m^2
6) Icm < I_right < I_left
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.