Question

The three 200 g masses are connected by massless, rigid rods. A. What is the triangle’s moment of inertia about the

axis through the center?

b. What is the triangle’s kinetic energy if it rotates

about the axis at 5. 0 rev/s?

Answer 1

The moment of inertia of the system can be calculated by summing up the moments of inertia of its individual components. The kinetic energy of the system can be calculated using the formula K = (1/2) × I × w^2.

Step-by-step explanation:

Moment of Inertia: The moment of inertia of a system can be calculated by summing up the moments of inertia of its individual components. In this case, the system consists of three 200 g masses connected by massless, rigid rods. The moment of inertia of each mass can be calculated using the formula I = m × r^2, where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation. Since all three masses are the same, their individual moments of inertia will be equal. Therefore, the moment of inertia of the system can be calculated as three times the moment of inertia of one mass.

The moment of inertia of one mass is given as I = 200 g × (r/2)^2. Since there are three masses, the moment of inertia of the system is 3 × 200 g × (r/2)^2.

Kinetic Energy:

The kinetic energy of a rotating object can be calculated using the formulaK = (1/2) × I × w^2. where K is the kinetic energy, I is the moment of inertia, and w is the angular velocity in radians per second. Substituting the moment of inertia of the system obtained earlier and the given angular velocity, we can calculate the kinetic energy of the system.