Certainly! Let's solve this problem step by step to calculate the rotational kinetic energy of the motorcycle wheel, starting with the moment of inertia.
### Step 1: Calculation of the Average Radius
Given the two radii [tex]\(R_1 = 0.290 \, \text{m}\)[/tex] and [tex]\(R_2 = 0.320 \, \text{m}\)[/tex]:
[tex]\[ \text{Average Radius} = \frac{R_1 + R_2}{2} = \frac{0.290 + 0.320}{2} = 0.305 \, \text{m} \][/tex]
### Step 2: Calculation of the Moment of Inertia [tex]\(I\)[/tex]
The moment of inertia for the wheel (assuming it can be approximated by a thin hoop of radius [tex]\( \text{Average Radius} \)[/tex] and mass [tex]\( m \)[/tex]) is given by:
[tex]\[ I = m \times (\text{Average Radius})^2 \][/tex]
Given the mass [tex]\( m = 12.5 \, \text{kg} \)[/tex]:
[tex]\[ I = 12.5 \times (0.305)^2 \][/tex]
To four significant figures, the moment of inertia is:
[tex]\[ I = 12.5 \times 0.093025 = 1.1628125 \approx 1.1628 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 3: Calculation of the Rotational Kinetic Energy [tex]\( KE_{\text{rot}} \)[/tex]
The formula for the rotational kinetic energy is:
[tex]\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \][/tex]
where [tex]\( \omega \)[/tex] is the angular velocity.
Given [tex]\( \omega = 120 \, \text{rad/s} \)[/tex] and [tex]\( I = 1.1628 \, \text{kg} \cdot \text{m}^2 \)[/tex]:
[tex]\[ KE_{\text{rot}} = \frac{1}{2} \times 1.1628 \times (120)^2 \][/tex]
Carrying out the calculation:
[tex]\[ KE_{\text{rot}} = 0.5 \times 1.1628 \times 14400 = 8371.68 \][/tex]
To one decimal place, the rotational kinetic energy is:
[tex]\[ KE_{\text{rot}} \approx 8372.2 \, \text{Joules} \][/tex]
### Final Answer:
- Moment of inertia, [tex]\( I \approx 1.1628 \, \text{kg} \cdot \text{m}^2 \)[/tex]
- Rotational kinetic energy, [tex]\( KE_{\text{rot}} \approx 8372.2 \, \text{J} \)[/tex]
So the filled blanks are:
Moment of inertia for the wheel:
[tex]\[ I = 1.1628 \, \text{kg} \cdot \text{m}^2 \][/tex]
Rotational kinetic energy:
[tex]\[ KE_{\text{rot}} = 8372.2 \, \text{Joules} \][/tex]
