Question

If the rotating mass of a rim type flywheel is distributed on a second rim type flywheel whose mean radius is half the mean radius of the former, then the energy stored in the second flywheel at the same speed will be ------- times that of the first flywheel

• A. \( \dfrac{1}{4} \)
• B. \( \dfrac{1}{2} \)
• C. \( 2 \)
• D. \( 4 \)

Hint:

For rim type flywheels at the same speed, energy stored is proportional to the square of radius.

Answer 1

The Correct Option is A

Step 1: Let the radius ratio between the second and first flywheel be \( k = \dfrac{r_2}{r_1} = \dfrac{1}{2} \), with mass and angular speed unchanged.
Step 2: Since kinetic energy of a rim-type flywheel is \( E = \tfrac{1}{2}mr^2\omega^2 \), the ratio of energies depends only on the square of the radius ratio:
\[ \dfrac{E_2}{E_1} = \left(\dfrac{r_2}{r_1}\right)^2 = k^2 \]
Step 3: Substituting \( k = \dfrac{1}{2} \):
\[ \dfrac{E_2}{E_1} = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4} \]

Answer 2

Plugging in concrete numbers makes the scaling easy to verify directly. Suppose the first flywheel has mass \( m = 8\,\text{kg} \), radius \( r = 2\,\text{m} \), and both flywheels spin at \( \omega = 1\,\text{rad/s} \) for simplicity. The second flywheel then has the same mass but radius \( 1\,\text{m} \). Testing each option against the computed energies:

1. \( \tfrac14 \): \( E_1 = \tfrac12(8)(2)^2(1)^2 = 16\,\text{J} \) and \( E_2 = \tfrac12(8)(1)^2(1)^2 = 4\,\text{J} \), giving \( \dfrac{E_2}{E_1} = \dfrac{4}{16} = \dfrac14 \), which matches this option exactly.
2. \( \tfrac12 \): This would require \( E_2 = 8\,\text{J} \), but the actual computed value is \( 4\,\text{J} \), so this option does not match the numeric result.
3. \( 2 \): This would require the second (smaller) flywheel to store more energy (\( 32\,\text{J} \)) than the first, which contradicts the direct calculation showing it stores less.
4. \( 4 \): This would require \( E_2 = 64\,\text{J} \), far more than the computed \( 4\,\text{J} \), and corresponds instead to doubling rather than halving the radius.

The direct numeric computation confirms the energy ratio is \( \tfrac14 \).

Therefore, the correct answer is \( \tfrac14 \).