A light, rigid rod of length / = 1.00 m joins two particles, with masses m1 = 4.

Question

A light, rigid rod of length / = 1.00 m joins two particles, with masses m1 = 4.00 kg and m2 3.00 kg, at its ends. The combination rotates in the xy plane about a pivot through the center of the rod (see figure below) Determine the angular momentum of the system about the origin when the speed of each particle is 2.50 m/s Part 1 of 3- Conceptualize This massive object is more like a weapon than a majorette's baton. We expect the angular momentum to be several ka m2/ Part 2 of 3- Categorize We compute the angular momentum for the system of two objects Part 3 of 3 - Analyze We write the equation for the total angular momentum of the system. Because the radius of the circular path for each object is 1/2, we have the following kg)(2.50 m/s)(| kg m2/s |m) + (| |kg)(2.50 m/-X - kg2.50 m/s Thus, in unit-vector notation, the angular momentum vector is given by k kg m/s

Explanation / Answer

moment of inertia
I = mr² = 4.00kg * (0.5m)² + 3.00kg * (0.5m)² = 1.75 kg·m²

= v / r = 2.50m/s / 0.50m = 5 rad/s

L = I = 1.75kg·m² * 5 rad/s = 8.75 kg·m²/s

The "right hand rule" tells us that the direction is along the +z axis.

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