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.
