A light, rigid rod l = 5.55 m in length rotates

in the xy plane about a pivot through the
rod’s center. Two particles of masses m1 =
9.8 kg and m2 = 2.4 kg are connected to its
ends.

Determine the angular momentum of the
system about the origin at the instant the
speed of each particle is v = 1.83 m/s.
Answer in units of kg m2
/s

To determine the angular momentum of the system about the origin, we need to use the principle of conservation of angular momentum. The angular momentum of a particle is given by the product of its moment of inertia and its angular velocity.

The moment of inertia for each particle can be calculated using the formula I = m * r^2, where m is the mass of the particle and r is the distance from the particle to the axis of rotation. Since the particles are connected to the ends of the rod that is rotating about its center, the distance for both particles will be half the length of the rod, which is l/2.

For particle 1:
m1 = 9.8 kg
r1 = l/2 = 5.55 m/2 = 2.775 m
I1 = m1 * r1^2 = 9.8 kg * (2.775 m)^2 = 74.169 kg m^2

For particle 2:
m2 = 2.4 kg
r2 = l/2 = 5.55 m/2 = 2.775 m
I2 = m2 * r2^2 = 2.4 kg * (2.775 m)^2 = 16.1744 kg m^2

Now, the angular momentum of each particle is given by L = I * ω, where ω is the angular velocity.

The angular velocity can be calculated using the formula v = ω * r, where v is the linear velocity and r is the distance from the particle to the axis of rotation.

For both particles, the linear velocity is v = 1.83 m/s, and the distance r is equal to l/2 = 2.775 m.

Using the formula ω = v / r, we can calculate the angular velocity for both particles:

For particle 1:
v1 = 1.83 m/s
r1 = 2.775 m
ω1 = v1 / r1 = 1.83 m/s / 2.775 m = 0.6595 rad/s

For particle 2:
v2 = 1.83 m/s
r2 = 2.775 m
ω2 = v2 / r2 = 1.83 m/s / 2.775 m = 0.6595 rad/s

Finally, we can calculate the total angular momentum of the system by adding the angular momentum of each particle:

L_total = L1 + L2 = I1 * ω1 + I2 * ω2 = (74.169 kg m^2)(0.6595 rad/s) + (16.1744 kg m^2)(0.6595 rad/s) = 81.755425 kg m^2/s

Therefore, the angular momentum of the system about the origin at the instant the speed of each particle is v = 1.83 m/s is 81.755425 kg m^2/s.