The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicualr to its plane is K. If the wheel is roling without slipping. Then the ratio of tis rotational kinetic energy to its translational kinetic energy is

Options

A. `(K^2)/(R^2)`
B. `(R^2)/(K^2)`
C. `(R^2)/(R^2 + K^2)`
D. `(K^2)/(R^2 + K^2)`

Answer

To solve the problem, we need to find the ratio of the rotational kinetic energy to the translational kinetic energy of a wheel that is rolling without slipping. Let's break down the solution step by step.

### Step 1: Define the Kinetic Energies
1. **Translational Kinetic Energy (TKE)**:
The translational kinetic energy of the wheel can be expressed as:
\[ TKE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the wheel and \( v \) is its linear velocity.

2. **Rotational Kinetic Energy (RKE)**:
The rotational kinetic energy of the wheel can be expressed as:
\[ RKE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the wheel and \( \omega \) is its angular velocity.

### Step 2: Relate Linear and Angular Velocity
Since the wheel is rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = R \omega \]
where \( R \) is the radius of the wheel.

### Step 3: Moment of Inertia
The moment of inertia \( I \) of the wheel about its axis is given by:
\[ I = mK^2 \]
where \( K \) is the radius of gyration.

### Step 4: Substitute into Kinetic Energy Equations
1. **Translational Kinetic Energy**:
Substituting \( v = R \omega \) into the translational kinetic energy formula:
\[ TKE = \frac{1}{2} m (R \omega)^2 = \frac{1}{2} m R^2 \omega^2 \]

2. **Rotational Kinetic Energy**:
Substituting \( I = mK^2 \) into the rotational kinetic energy formula:
\[ RKE = \frac{1}{2} (mK^2) \omega^2 = \frac{1}{2} m K^2 \omega^2 \]

### Step 5: Calculate the Ratio of RKE to TKE
Now, we can find the ratio of the rotational kinetic energy to the translational kinetic energy:
\[ \text{Ratio} = \frac{RKE}{TKE} = \frac{\frac{1}{2} m K^2 \omega^2}{\frac{1}{2} m R^2 \omega^2} \]
The \( \frac{1}{2} m \) and \( \omega^2 \) terms cancel out:
\[ \text{Ratio} = \frac{K^2}{R^2} \]

### Final Answer
Thus, the ratio of the rotational kinetic energy to the translational kinetic energy is:
\[ \frac{K^2}{R^2} \]