A uniform rod (length = 2.9m) of negligible mass has a 3kg point mass A attached

Question

A uniform rod (length = 2.9m) of negligible mass has a 3kg point mass A attached to one end and a 5kg point mass B attached to the other end. The rod is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through a point 1.359m from mass B. The rod is released from rest when it is horizontal. a) What is the moment of inertia of the system? kg m^2 b) What is the change in potential energy of the system from initial to final position when mass B is in the lowest position? (sign is important) J c) Find the angular speed omega when mass B passes through the lowest point, rad/s

Explanation / Answer

(a)

I_tot = I_A + I_B = mA rA^2 + mB rB^2 = ( 3)(2.9-1.359)^2 + ( 5 kg ( 1.359)^2 = 16.35 kg m^2

(b)

U_f - mA g yA + mB g yB = 3(9.8)( 2.9-1.359)+5(9.8)(-1.359) = -21.28 J

del U = Uf- Ui = -21.28 J - 0 = -21.28 J

(c)

Rf= Ei

Kf + Uf = Ki + Ui

1/2 I w^2 + (-21.28) = 0

1/2 I w^2 = 21.2856 J

w = sqrt 2( 21.2856)/I total

= sqrt 2 ( 21.2856)/16.35

=1.61 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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