The speed of a moving car is `60kmh^(-1)`. The wheels having diameter of 0.60 m are stopped in 10 rotations by applying brakes. What will be the angular retardation produced by brakes?

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the speed from km/h to m/s
The speed of the car is given as 60 km/h. To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \text{ km/h} = \frac{5}{18} \text{ m/s} \]
Thus,
\[ \text{Speed in m/s} = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \text{ m/s} \approx 16.67 \text{ m/s} \]

### Step 2: Calculate the radius of the wheel
The diameter of the wheel is given as 0.60 m. The radius \( R \) is half of the diameter:
\[ R = \frac{0.60}{2} = 0.30 \text{ m} \]

### Step 3: Calculate the initial angular velocity \( \omega \)
Using the relationship between linear velocity \( V \) and angular velocity \( \omega \):
\[ V = R \omega \]
We can rearrange this to find \( \omega \):
\[ \omega = \frac{V}{R} = \frac{\frac{50}{3}}{0.30} = \frac{50}{3} \times \frac{10}{3} = \frac{500}{9} \text{ rad/s} \]

### Step 4: Calculate the angular displacement \( \theta \)
The wheel stops after 10 rotations. The angular displacement in radians for 10 rotations is given by:
\[ \theta = 10 \times 2\pi = 20\pi \text{ radians} \]

### Step 5: Apply the equation of motion for angular motion
We know that:
\[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]
Where:
- \( \omega_f \) is the final angular velocity (0 rad/s, since the wheel stops),
- \( \omega_i \) is the initial angular velocity (\( \frac{500}{9} \) rad/s),
- \( \alpha \) is the angular retardation (which we need to find),
- \( \theta \) is the angular displacement (20π radians).

Substituting the known values:
\[ 0 = \left(\frac{500}{9}\right)^2 + 2\alpha (20\pi) \]

### Step 6: Solve for angular retardation \( \alpha \)
Rearranging the equation gives:
\[ 2\alpha (20\pi) = -\left(\frac{500}{9}\right)^2 \]
\[ \alpha = -\frac{\left(\frac{500}{9}\right)^2}{40\pi} \]
Calculating \( \left(\frac{500}{9}\right)^2 \):
\[ \left(\frac{500}{9}\right)^2 = \frac{250000}{81} \]
Now substituting back:
\[ \alpha = -\frac{\frac{250000}{81}}{40\pi} = -\frac{250000}{3240\pi} \]
Calculating this gives:
\[ \alpha \approx -24.56 \text{ rad/s}^2 \]

### Final Result:
The angular retardation produced by the brakes is approximately \( -24.56 \text{ rad/s}^2 \).

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