Two small 0.55-kg masses are attached to opposite ends of a very lightweight rigid rod 0.3 m long. The system is spinning in a horizontal plane around a vertical axis perpendicular to the rod located halfway between the masses. Each mass is moving in a circle of radius 0.15 m at a speed of 0.90 m/s. What is the total angular momentum of this system?
Angular momentum is the rotational equivalent of linear momentum. It depends on the moment of inertia (I), which is a measure of an object's resistance to changes in its rotational motion, and the angular velocity (ω), which is the rate at which an object rotates.
To calculate the total angular momentum of the system, we need to calculate the individual angular momentum of each mass and then add them together.
The formula for angular momentum is L = I * ω, where L is the angular momentum.
First, let's calculate the moment of inertia for each mass. For a point mass (like these masses attached to a rod), the moment of inertia is given by I = m * r^2, where m is the mass and r is the radius.
Given:
Mass (m) = 0.55 kg
Radius (r) = 0.15 m
Calculating moment of inertia for each mass:
I = m * r^2 = 0.55 kg * (0.15 m)^2 = 0.012375 kg·m^2
Now, let's calculate the angular velocity (ω). The angular velocity is given by ω = v / r, where v is the linear velocity and r is the radius.
Given:
Linear Velocity (v) = 0.90 m/s
Radius (r) = 0.15 m
Calculating angular velocity for each mass:
ω = v / r = 0.90 m/s / 0.15 m = 6 rad/s
Now, let's calculate the angular momentum for each mass:
L = I * ω = 0.012375 kg·m^2 * 6 rad/s = 0.07425 kg·m^2/s
Since there are two masses, we need to add their angular momenta together:
Total Angular Momentum = 2 * L = 2 * 0.07425 kg·m^2/s = 0.1485 kg·m^2/s
Therefore, the total angular momentum of the system is 0.1485 kg·m^2/s.