© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Convert the linear speed into m/s
The automobile’s speed is given as 54 km/h. We convert this into m/s using the relation
$1 \,\text{km/h} = \displaystyle \frac{5}{18}\,\text{m/s}$:
$ v = 54 \times \frac{5}{18} \,\text{m/s} = 54 \times 0.277\ldots \,\text{m/s} = 15 \,\text{m/s}$
Step 2: Identify given parameters
Linear speed (now in m/s): $v = 15 \,\text{m/s}$
Wheel radius: $R = 0.45 \,\text{m}$
Moment of inertia of the wheel: $I = 3 \,\text{kg}\,\text{m}^2$
Time taken to come to rest: $t = 15 \,\text{s}$
Step 3: Calculate the initial angular speed of the wheel
Since $ \omega_i = \displaystyle \frac{v}{R} $, we have
$ \omega_i = \displaystyle \frac{15 \,\text{m/s}}{0.45 \,\text{m}} = \frac{15}{0.45} \,\text{rad/s} = \frac{1500}{45} \,\text{rad/s} = \frac{100}{3} \,\text{rad/s}.$
Step 4: Determine the final angular speed
When the automobile comes to rest, the final angular speed of its wheels is
$ \omega_f = 0 \,\text{rad/s}.$
Step 5: Compute the angular acceleration (or deceleration)
The angular deceleration $ \alpha $ is given by
$ \alpha = \displaystyle \frac{\omega_f - \omega_i}{t}
= \frac{0 - \frac{100}{3}}{15} \,\text{rad/s}^2
= - \frac{100}{45} \,\text{rad/s}^2.$
Here, the negative sign indicates a decrease in angular velocity (retardation). We will use the magnitude for torque calculation.
Step 6: Calculate the required braking torque
Torque $ \tau $ can be found using the relation
$ \tau = I \times |\alpha| = 3 \,\text{kg}\,\text{m}^2 \times \frac{100}{45} \,\text{rad/s}^2.$
Hence,
$ \tau = \frac{300}{45} \,\text{kg}\,\text{m}^2 \,\text{s}^{-2}
= \frac{20}{3} \,\text{kg}\,\text{m}^2 \,\text{s}^{-2}
= 6.66 \,\text{kg}\,\text{m}^2 \,\text{s}^{-2}.$
Final Answer
The magnitude of the average torque transmitted by the brakes to each wheel is
$6.66 \,\text{kg}\,\text{m}^2 \,\text{s}^{-2}.$
