What is the moment of inertia of the object about an axis at the left end of the

Question

What is the moment of inertia of the object about an axis at the left end of the rod? 2) If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 434.0 N is exerted perpendicular to the rod at the center of the rod? 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) If the object is fixed at the center of mass, what is the angular acceleration if a force F = 434.0 N is exerted parallel to the rod at the end of rod? 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

m1=7,44 kg L=5,04 m m2=37,2 kg R=1,26 m sqr(x) means x*x (1) The moment of inertia for a rod rotating around its center is J1=1/12*m*sqr(r) In this case J1=1/12*m1*sqr(L) J1=15,749 kg*m2 The moment of inertia for a solid sphere rotating around its center is J2=2/5*m*sqr(r) In this case J2=2/5*m2*sqr(R) J2=23,623 kg*m2 As the thigy rotates around the free end of the rod then for the sphere the axis around what it rotates is at a distance of d2=L+R For the rod it is d1=1/2*L From Steiner theorem for the rod we get J1"=J1+m1* sqr(d1) J1"=62,996 kg*m2 for the sphere we get J2"=J2+m2*sqr(d2) J2"=1500,091 kg*m2 And the total moment of inertia for the first case is Jt1=J1"+J2" FIRST ANSWER Jt1=1563,087 kg*m2 (2) F=488 N The torque given to a system in general is M=F*d*sin(a) where a is the angle between F and d and where d is the distance from the rotating axis. In this case a=90" and so M=F*L/2 M=1229,76 Nm The acceleration can be found from e1=M/Jt1 SECOND ANSWER e1=0,787 rad/s2 (3) I assume the text to be right in the case where the center of mass is. Again we have to use Steiner theorem In this case h1=(L+R)/2 and h2=R/2 So J1""=J1+m1*sqr(h1) J1""=89,572 kg*m2 J2""=J2+m2*sqr(h2) J2""=82,682 kg*m2 and Jt2=J1""+J2"" THIRD ANSWER Jt2=172,254 kg*m2 (4) F=488 N M=F*(L+R/2)*sin(a) In this case a=0" and so M=0 and thus FOURTH ANSWER e2=0 rad/s2 (5) In this case again we have to use Steiner theorem k1=2*R+L/2 K2=R so J1"""=J1+m1*sqr(k1) J1"""=204,737 kg*m2 J2"""=J2+m2*sqr(k2) J2"""=82,682 kg*m2 Jt3=J1"""+J2""" So THE FINAL ANSWER Jt3=287,419 kg*m2 similar kind of problem.

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