Question

A system of point particles is rotating about a fixed axis at 4 rev/s. The particles are fixed with respect to each other. The masses and distances to the axis of the point particles are m1=0.1kg,r1=0.2m, m2=0.05kg,r2=0.4m, m3=0.5kg,r3=0.01m. (a) What is the moment of inertia of the system? (b) What is the rotational kinetic energy of the system?

(a) 0.035 kg⋅m²
(b) 0.040 kg⋅m²
(c) 0.045 kg⋅m²
(d) 0.050 kg⋅m²

Answer 1

The moment of inertia of the system is 0.035 kg⋅m² and the rotational kinetic energy is 0.040 kg⋅m².

Step-by-step explanation:

The moment of inertia of the system can be calculated by using the formula Σmjr², where Σmjr² represents the sum of the masses of the particles multiplied by the square of their distances from the axis of rotation. In this case, the moment of inertia is given by:

I = (0.1 kg)(0.2 m)² + (0.05 kg)(0.4 m)² + (0.5 kg)(0.01 m)² = 0.035 kg⋅m²

The rotational kinetic energy of the system can be calculated using the formula ½Iω², where I is the moment of inertia and ω is the angular velocity. In this case, the rotational kinetic energy is:

K = ½(0.035 kg⋅m²)(2π(4 rev/s))^2 = 0.040 kg⋅m²