Question

A wheel is rotating freely with an angular speed $\omega$ on a shaft. The moment of inertia of the wheel is $I$ and the moment of inertia of the shaft is negligible. Another wheel of the moment of inertia $3I$ initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is:

• A. $\frac{3}{4}$
• B. $0$
• C. $\frac{5}{6}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A

$\frac{3}{4}$

Step 1. Given data

The angular speed of the wheel is $\omega$.

The moment of inertia of the wheel is $I$

Another wheel of the moment of inertia is $3I$

Step 2. Calculating the angular speed of another wheel

According to the angular momentum conservation,

Initial angular momentum $=$ Final angular momentum

$$\begin{gathered} I\omega =I\omega \text{'}+3I\omega \text{'} \\ I\omega =4I\omega \text{'} \\ \omega \text{'}=\frac{\omega }{4} \end{gathered}$$ [ Momentum is the product of the moment of inertia and the angular speed.]

Where, $I$ is the moment of inertia.

$\omega$ is the angular speed

$\omega \text{'}$ is the final angular speed

Step 3. Finding the initial and final kinetic energy,

Kinetic energy of rotational motion is equal to the half of the product of the moment of inertia and the square of the angular velocity.

Initial kinetic energy, ${(K.E.)}_i=\frac{1}{2}I{\omega }^2$

Final kinetic energy, ${(K.E.)}_f=\frac{1}{2}\left(I+3I\right)\omega {\text{'}}^2$

Step 4. Calculating loss in the Kinetic energy

The kinetic energy loss is given as

Loss in kinetic energy, ${\left(∆K.E.\right)}_l$ $=$ Initial kinetic energy, ${\left(∆K.E.\right)}_i$ $-$Final kinetic energy, ${\left(∆K.E.\right)}_f$

$$\begin{gathered} {\left(∆K.E.\right)}_l=\frac{1}{2}I{\omega }^2-\frac{1}{2}\left(I+3I\right){\left(\frac{\omega }{4}\right)}^2 \\ {\left(∆K.E.\right)}_l=\frac{1}{2}I{\omega }^2-\frac{1}{2}\left(4I\right){\left(\frac{\omega }{4}\right)}^2 \\ {\left(∆K.E.\right)}_l=\frac{1}{2}I{\omega }^2\left(1-\frac{1}{4}\right) \\ {\left(∆K.E.\right)}_l=\frac{3}{8}I{\omega }^2 \end{gathered}$$ $$\begin{gathered} \left[{\left(∆K.E.\right)}_i=\frac{1}{2}I{\omega }^2 \\ {\left(∆K.E.\right)}_f=\frac{1}{2}\left(I+3I\right)\omega {\text{'}}^2\right] \\ \left[\omega \text{'}=\frac{\omega }{4}\right] \end{gathered}$$

Therefore, the fractional loss is given as

$$\begin{gathered} =\frac{{\left(∆K.E.\right)}_l}{\left(∆K.E.\right)i} \\ =\frac{{\displaystyle \frac{3}{8}}I{\omega }^2}{{\displaystyle \frac{1}{2}}I{\omega }^2} \\ =\frac{3}{4} \end{gathered}$$

Hence, the correct option is (A).