An autmobile moves on road with a speed of `54 km//h`. The radius of its wheel is `0.45 m` and the moment of inertia of the wheel about its axis of rotation is `3 kg m^(2)`. If the vehicle is brought to rest in `15 s`, the magnitude of average torque tansmitted by its brakes to the wheel is :

Options

A. `6.66 kg m^(2) s^(-2)`
B. `8.58 kg m^(2) s^(-2)`
C. `10.86 kg m^(2) s^(-2)`
D. `2.86 kg m^(2) s^(-2)`

Answer

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Convert the speed from km/h to m/s
Given speed of the automobile is \(54 \, \text{km/h}\).

To convert km/h to m/s, we use the conversion factor:
\[ 1 \, \text{km/h} = \frac{5}{18} \, \text{m/s} \]
Thus,
\[ 54 \, \text{km/h} = 54 \times \frac{5}{18} = 15 \, \text{m/s} \]

### Step 2: Calculate the initial angular velocity (\(\omega_i\))
The initial linear speed \(V\) is related to the angular velocity \(\omega\) by the equation:
\[ \omega = \frac{V}{R} \]
where \(R\) is the radius of the wheel.

Given \(R = 0.45 \, \text{m}\), we find:
\[ \omega_i = \frac{15 \, \text{m/s}}{0.45 \, \text{m}} = \frac{15}{0.45} \, \text{rad/s} = 33.33 \, \text{rad/s} \]

### Step 3: Determine the final angular velocity (\(\omega_f\))
Since the vehicle comes to rest, the final angular velocity is:
\[ \omega_f = 0 \, \text{rad/s} \]

### Step 4: Calculate the angular acceleration (\(\alpha\))
Angular acceleration can be calculated using the formula:
\[ \alpha = \frac{\omega_f - \omega_i}{t} \]
where \(t\) is the time taken to stop, given as \(15 \, \text{s}\).

Substituting the values:
\[ \alpha = \frac{0 - 33.33}{15} = -2.22 \, \text{rad/s}^2 \]

### Step 5: Calculate the average torque (\(\tau\))
The torque is related to angular acceleration by the equation:
\[ \tau = I \cdot \alpha \]
where \(I\) is the moment of inertia of the wheel, given as \(3 \, \text{kg m}^2\).

Substituting the values:
\[ \tau = 3 \, \text{kg m}^2 \cdot (-2.22 \, \text{rad/s}^2) = -6.66 \, \text{N m} \]

Since we are interested in the magnitude of the torque, we take the absolute value:
\[ |\tau| = 6.66 \, \text{N m} \]

### Final Answer
The magnitude of the average torque transmitted by the brakes to the wheel is:
\[ \boxed{6.66 \, \text{N m}} \]

---