Three 0.50 kg particles form an equilateral triangle with 0.60 msides. The parti

Question

Three 0.50 kg particles form an equilateral triangle with 0.60 msides. The particles are connected by rods of negligible mass. Whatis the rotational inertia of this rigid body about(a) an axis that passes through one of theparticles and is parallel to the rod connecting the other two,(b) an axis that passes through the midpoint ofone of the sides and is perpendicular to the plane of the triangle,and (c) an axis that is parallel to one side ofthe triangle and passes through the midpoints of the other twosides?

Explanation / Answer

(a) One particle is onthe axis, so r= 0 for it. For each of the others,
   the distance from the axis is r =(0.60m)sin60o = 0.52m     Therefore, the rotational inertiais I =miri2
        
                                                        = 0.27 kg.m2


(b) The two particles that are nearest the axis are each adistance of r= 0.30 m fromit.
  The particle “opposite” from that side is adistance r= (0.60 m) sin 60° = 0.52 m fromthe
   axis. Thus, the rotational inertia isI=miri2
                                                      = 0.22 kg .m2

(c) The distance fromthe axis for each of the particles is r=1/2(0.60m) sin600 . Now,
                                I = 2(5.0 kg)(0.26m)2
Simplifying we get theanswer.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.