Calculate the rotational rotational inertia of a wheel that has a kinetic energy of 24,400 J when rotating at 602 rev/min.
The Correct Answer and Explanation is:
To calculate the rotational inertia (moment of inertia) of the wheel, we can use the formula for rotational kinetic energy:KE=12Iω2KE = \frac{1}{2} I \omega^2KE=21Iω2
Where:
- KEKEKE is the kinetic energy,
- III is the moment of inertia,
- ω\omegaω is the angular velocity in radians per second.
We are given:
- KE=24,400 JKE = 24,400 \, \text{J}KE=24,400J,
- rev/min=602 rev/min\text{rev/min} = 602 \, \text{rev/min}rev/min=602rev/min.
Step 1: Convert the angular velocity from rev/min to rad/s.
To convert the angular velocity from revolutions per minute to radians per second, use the following conversion factors:
- 1 rev=2π radians1 \, \text{rev} = 2\pi \, \text{radians}1rev=2πradians,
- 1 min=60 seconds1 \, \text{min} = 60 \, \text{seconds}1min=60seconds.
Thus, the angular velocity ω\omegaω in radians per second is:ω=602 rev/min×2π rad1 rev×1 min60 seconds=62.97 rad/s\omega = 602 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{seconds}} = 62.97 \, \text{rad/s}ω=602rev/min×1rev2πrad×60seconds1min=62.97rad/s
Step 2: Rearrange the formula for moment of inertia.
Now that we know KE=24,400 JKE = 24,400 \, \text{J}KE=24,400J and ω=62.97 rad/s\omega = 62.97 \, \text{rad/s}ω=62.97rad/s, we can rearrange the equation to solve for III:I=2KEω2I = \frac{2KE}{\omega^2}I=ω22KE
Step 3: Substitute the known values.
Substitute the values for kinetic energy and angular velocity:I=2×24,400(62.97)2I = \frac{2 \times 24,400}{(62.97)^2}I=(62.97)22×24,400I=48,8003965.1≈12.3 kg⋅m2I = \frac{48,800}{3965.1} \approx 12.3 \, \text{kg} \cdot \text{m}^2I=3965.148,800≈12.3kg⋅m2
Final Answer:
The rotational inertia of the wheel is approximately 12.3 kg·m².
This calculation assumes that the wheel’s mass is uniformly distributed, and that the moment of inertia depends only on the wheel’s angular velocity and kinetic energy. If the shape or mass distribution of the wheel were more specific (e.g., a solid disk or hoop), the formula for III would change accordingly.
