Topic: Theory of Machines
If the rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half the mean radius of the former, then energy stored in the later at the same speed will be.
- A. Four times the first one
- B. Same as the first one
- C. One fourth of the first one
- D. One and a half times the first one
Answer: C. One fourth of the first one
Explanation: Energy stored in a flywheel is $E = \frac{1}{2}I\omega^{2}$. For a rim type flywheel, $I = mR^{2}$. Thus, $E_{1} = \frac{1}{2}mR_{1}^{2}\omega^{2}$. When the radius is half the mean radius ($R_{2} = R_{1}/2$), the energy is $E_{2} = \frac{1}{2}m(R_{1}/2)^{2}\omega^{2} = \frac{1}{4} \times \frac{1}{2}mR_{1}^{2}\omega^{2} = \frac{1}{4}E_{1}$.