Question

4) a system of point particles is shown in the following figure. each particle has mass 0.3 kg and they all lie in the same plane. (a) what is the rotational inertia of the system about the given axis? (b) if the system rotates at 5.00 rev/s, what is its rotational kinetic energy?

Answer 1

The moment of inertia of the system is 0.0035 kg · m² and its rotational kinetic energy is approximately 2.19 J.

Step-by-step explanation:

The moment of inertia of the system about the given axis can be calculated by using the definition of moment of inertia for a system of particles. We can pull the mass of each particle in front of the summation symbol and evaluate the quantity. In this case, the moment of inertia is equal to 0.0035 kg · m².

The rotational kinetic energy of the system can be calculated using the formula: rotational kinetic energy = (1/2) * I * ω², where I is the moment of inertia and ω is the angular velocity. Substituting the given values into the formula, the rotational kinetic energy of the system is approximately 2.19 J.

Answer 2

The rotational inertia of the system is 0.168 kg⋅m².

The rotational kinetic energy of the system is approximately 84.0π J.

Part (a): Rotational Inertia

To find the rotational inertia of the system about the given axis, we can use the formula:

I = Σmr²

where:

I is the rotational inertia

m is the mass of each particle (0.3 kg in this case)

r is the distance of each particle from the axis of rotation

From the image, we can see that the distances of the particles from the axis are:

Particle A: 0.2 m

Particle B: 0.4 m

Particle C: 0.6 m

Therefore, the rotational inertia is:

I = 3 × 0.3 kg × [(0.2 m)² + (0.4 m)² + (0.6 m)²]

I = 0.3 kg × (0.04 + 0.16 + 0.36)

I = 0.3 kg × 0.56

I = 0.168 kg⋅m²

Part (b): Rotational Kinetic Energy

The rotational kinetic energy of the system can be calculated using the formula:

KE = ½ × I × ω²

where:

KE is the rotational kinetic energy

I is the rotational inertia (0.168 kg⋅m² in this case)

ω is the angular velocity (5.00 rev/s)

We need to convert the angular velocity from revolutions per second to radians per second:

ω = 5.00 rev/s × 2π rad/rev

ω = 10.0π rad/s

Now we can calculate the rotational kinetic energy:

KE = ½ × 0.168 kg⋅m² × (10.0π rad/s)²

KE = 0.5 × 0.168 kg⋅m² × 100π² rad²/s²

KE ≈ 84.0π J

Therefore, the rotational kinetic energy of the system is approximately 84.0π J.

The missing image from question: