A 1.90-kg particle moves in the xy plane with a velocity of v rightarrow = (4.40

Question

A 1.90-kg particle moves in the xy plane with a velocity of v rightarrow = (4.40 i^ ? 3.70 j^ ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r rightarrow = (1.50 i^+ 2.20 j^) m. The direction of the angular momentum is defined to be perpendicular to the plane of rotation. i^ The direction of the angular momentum is defined to be perpendicular to the plane of rotation. j^ Review the equation for the angular momentum in terms of components of the vectors. K^) kg.m2/s

Explanation / Answer

Angular momentum is L = Iw = kmr^2 w = km r X v = m (r X v) = m (1.6 i + 2.2j) X (4.6 i - 3.1j) = m (1.6*4.6 i X i - 1.6 * 3.1 i X j + 2.2 *4.6 j X i - 2.2* 3.1 j X j) = m (0 - 4.9 k + 10.12 (-k) - 0) = - m 15.02 k = - 1.70*15.02 k = -25.534 k ANS.

Or using all the axes L = 0 i + 0 j - 25.5 k ANS.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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