A strap is passing over a wheel of radius 30 cm. During the time the wheel moving with initial constant velocity of 2 rev/sec. comes to rest the strap covers a distance of 25 m. The deceleration of the wheel in rad/s is

Options

A. `0.94`
B. `1.2`
C. `2.0`
D. `2.5`

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the radius to meters
Given that the radius \( R = 30 \) cm, we need to convert this to meters:
\[ R = 30 \, \text{cm} = 0.30 \, \text{m} \]

### Step 2: Calculate the initial angular velocity in radians per second
The initial velocity of the wheel is given as \( 2 \) revolutions per second. To convert this to radians per second, we use the conversion factor \( 2\pi \) radians per revolution:
\[ \omega_i = 2 \, \text{rev/sec} \times 2\pi \, \text{rad/rev} = 4\pi \, \text{rad/sec} \]

### Step 3: Calculate the distance covered by the strap in terms of revolutions
The strap covers a distance of \( 25 \) m. The distance covered by the strap in one revolution is equal to the circumference of the wheel:
\[ \text{Circumference} = 2\pi R = 2\pi \times 0.30 \, \text{m} = 0.60\pi \, \text{m} \]
Let \( n \) be the number of revolutions. The total distance covered by the strap can be expressed as:
\[ n \times \text{Circumference} = 25 \, \text{m} \]
Thus,
\[ n \times 0.60\pi = 25 \]
Solving for \( n \):
\[ n = \frac{25}{0.60\pi} \approx \frac{25}{1.884} \approx 13.25 \, \text{revolutions} \]

### Step 4: Calculate the angular displacement in radians
The angular displacement \( \theta \) in radians for \( n \) revolutions is:
\[ \theta = n \times 2\pi = 13.25 \times 2\pi \approx 83.24 \, \text{radians} \]

### Step 5: Use the kinematic equation for angular motion
We know that the final angular velocity \( \omega_f = 0 \) (the wheel comes to rest). Using the equation:
\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]
Substituting the known values:
\[ 0 = (4\pi)^2 + 2\alpha(83.24) \]
This simplifies to:
\[ 0 = 16\pi^2 + 166.48\alpha \]
Solving for \( \alpha \):
\[ 166.48\alpha = -16\pi^2 \]
\[ \alpha = -\frac{16\pi^2}{166.48} \]

### Step 6: Calculate the deceleration
Now substituting \( \pi \approx 3.14 \):
\[ \alpha \approx -\frac{16 \times (3.14)^2}{166.48} \approx -\frac{16 \times 9.8596}{166.48} \approx -\frac{157.7536}{166.48} \approx -0.947 \, \text{rad/s}^2 \]
Thus, the deceleration of the wheel is approximately:
\[ \alpha \approx 0.94 \, \text{rad/s}^2 \]

### Final Answer
The deceleration of the wheel is approximately \( 0.94 \, \text{rad/s}^2 \).

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