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A light, rigid rod of length l = 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 5.80 m/s. magnitude kg A?· m2/s direction ---Select--- +x Acˆ’x +y Acˆ’y +z Acˆ’z

A light, rigid rod of length l = 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 5.80 m/s. magnitude kg A?· m2/s direction ---Select--- +x Acˆ’x +y Acˆ’y +z Acˆ’z

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A light, rigid rod of length l = 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 5.80 m/s. magnitude kg A?· m2/s direction ---Select--- +x Acˆ’x +y Acˆ’y +z Acˆ’z
Expert Solution
Step 1

Given:

  • The length of rod is l=1.00 m.
  • The mass of first particle is m1=4.00 kg.
  • The mass of second particle is m2=3.00 kg.
  • The speed of each particle is v=5.80 m/s.

The formula to calculate the distance of each particle from center of rod is,

r=l2

Here, r is the distance of each particle and l is the length.

Substitute the known value in the formula to calculate the distance of each particle.

r=1.00 m2=0.50 m

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