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Part 1- A flywheel in the form of a heavy circular disk of diameter 0.333 m and mass of 137 kg is mounted on a frictionless bearing. A motor connected to the flywheel accelerates it from rest to 1550 rev/min. What is the moment of inertia of the flywheel?

Part 2- How much work is done on it during this acceleration?

Part 3- After 1550 rev/min is achieved, the motor is disengaged. A friction brake is used to slow the rotational rate to 1118 rev/min. What is the magnitude of the energy dissipated as heat from the friction brake?

Respuesta :

MathPhys

Answer:

1. 1.90 kg m²

2. 25.0 kJ

3. 12.0 kJ

Explanation:

Convert rev/min to rad/s:

1550 rev/min × (2π rad/rev) × (1 min / 60 s) = 162.3 rad/s

1118 rev/min × (2π rad/rev) × (1 min / 60 s) = 117.1 rad/s

Part 1: The moment of inertia of a solid disk is:

I = ½ mr²

I = ½ (137 kg) (0.333 m / 2)²

I = 1.90 kg m²

Part 2: Work = change in energy

W = RE

W = ½ Iω²

W = ½ (1.90 kg m²) (162.3 rad/s)²

W = 25,000 J

W = 25.0 kJ

Part 3: Energy is conserved, so heat = change in energy

Q = ΔRE

Q = ½ I (ω² – ω₀²)

Q = ½ (1.90 kg m²) ((162.3 rad/s)² – (117.1 rad/s)²)

Q = 12,000 J

Q = 12.0 kJ

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