A uniform horizontal rod of mass M and length l rotates with angular velocity om

Question

A uniform horizontal rod of mass M and length l rotates with angular velocity omega about a vertical axis through its center attached to each end of the rod is a small mass m. Determine the angular momentum of the system about the axis, expressing your answer in terms of the variables M, m, l(lower case L, length of the rod) and omega. So I went about it by saying momentum final= momentum initial (L1= L2). Thus 1/12Ml^2 *omega = (1/12MI^2 +2ml^2) omega 2 but I'm unsure as to how to properly solve for angular momentum now, help will be greatly appreciated!

Explanation / Answer

I=M*l^2/12 +2*m*l^2/4 L=I*w=(M/12+m/2)*l^2*w

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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