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A bicycle wheel rotates at an angular speed of 13 rad/s. What is the
wheel's rotational kinetic energy if it has a mass of 2.80 kg and a radius of
0.300 m?



Answer :

To calculate the rotational kinetic energy of a bicycle wheel, we can use the formula: \[ KE_{rotation} = \frac{1}{2} I \omega^2 \] where \( KE_{rotation} \) is the rotational kinetic energy, \( I \) is the moment of inertia of the wheel, and \( \omega \) is the angular speed. For a solid cylinder-like shape, such as a bicycle wheel, the moment of inertia \( I \) is defined as: \[ I = \frac{1}{2} m r^2 \] Here, \( m \) is the mass of the wheel, and \( r \) is the radius of the wheel. Given: - The angular speed \( \omega = 13 \) rad/s - The mass of the wheel \( m = 2.80 \) kg - The radius of the wheel \( r = 0.300 \) m First, calculate the moment of inertia of the wheel: \[ I = \frac{1}{2} m r^2 \] \[ I = \frac{1}{2} \times 2.80 \text{ kg} \times (0.300 \text{ m})^2 \] \[ I = \frac{1}{2} \times 2.80 \times 0.09 \] \[ I = 1.4 \times 0.09 \] \[ I = 0.126 \text{ kg m}^2 \] Now that we have the moment of inertia, we can calculate the rotational kinetic energy: \[ KE_{rotation} = \frac{1}{2} I \omega^2 \] \[ KE_{rotation} = \frac{1}{2} \times 0.126 \text{ kg m}^2 \times (13 \text{ rad/s})^2 \] \[ KE_{rotation} = 0.063 \times 169 \] \[ KE_{rotation} = 10.647 \text{ J} \] So, the wheel's rotational kinetic energy is 10.647 joules.