What is the approximate moment of inertia of a bicycle wheel that weighs 19.6 N and has a diameter of 0.70 m?
The correct answer and explanation is:
The moment of inertia (I) of a bicycle wheel can be estimated using the formula for a solid disc, which is a good approximation for the wheel’s structure. The formula for the moment of inertia of a solid disc rotating about an axis through its center is: I=12mr2I = \frac{1}{2} m r^2
Where:
- mm is the mass of the wheel
- rr is the radius of the wheel
Step 1: Calculate the Mass of the Wheel
The weight of the bicycle wheel is given as 19.6 N. Weight (WW) is related to mass (mm) by the equation: W=mgW = m g
Where:
- gg is the acceleration due to gravity, approximately 9.8 m/s².
Rearranging to find mass: m=Wg=19.69.8=2 kgm = \frac{W}{g} = \frac{19.6}{9.8} = 2 \, \text{kg}
Step 2: Find the Radius
The diameter of the wheel is given as 0.70 m, so the radius rr is half of that: r=0.702=0.35 mr = \frac{0.70}{2} = 0.35 \, \text{m}
Step 3: Calculate the Moment of Inertia
Now, we can substitute the mass and radius into the moment of inertia formula: I=12×2×(0.35)2=1×0.1225=0.1225 kg⋅m2I = \frac{1}{2} \times 2 \times (0.35)^2 = 1 \times 0.1225 = 0.1225 \, \text{kg} \cdot \text{m}^2
Conclusion:
The approximate moment of inertia of the bicycle wheel is 0.1225 kg·m². This calculation assumes that the wheel is a solid disc, which is a reasonable approximation for most bicycle wheels, though some wheels may have spokes or other components that could slightly alter the actual moment of inertia.