The two 2.5-kg spheres are released from rest and gently nudged outward from the

Question

The two 2.5-kg spheres are released from rest and gently nudged outward from the position theta = 0 and then rotate in a vertical plane about the fixed centers of their attached gears, thus maintaining the same angle theta for both rods. Determine the velocity v of each sphere as the rods pass the position theta = 65 degree. The spring is unstretched when theta = 0, and the masses of the two identical rods and the two gear wheels may be neglected.

Explanation / Answer

initial P.E=2*2.5*10*470*10^-3=23.5 J final P.E=2*2.5*10*470Cos(65)*10^-3=9.9315 J change in P.E=13.56 This energy is used for increasing K.E and spring Energy elongation in spring x=2*235 Sin(65)=425.96 mm now spring energy=0.5*129*(425.96*10^-3)^2=11.703 J ----->K.E=13.56-11.703=1.8567=2*0.5*2.5*v^2------->v^2=0.74269----->v=0.86179 m/sec

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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