A stationary wheel starts rotating about its own axis at an angular acceleration `5.5 rad//s^(2)` . To acquire an angular velocity 420 revolutions per minute , the number of rotations made by the wheel is

Options

A. 14
B. 21
C. 28
D. 35

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the final angular velocity from revolutions per minute (rpm) to radians per second (rad/s).

Given:
- Final angular velocity, \( \omega_f = 420 \) revolutions per minute.

To convert revolutions per minute to radians per second, we use the conversion factor:
\[ 1 \text{ revolution} = 2\pi \text{ radians} \]
\[ 1 \text{ minute} = 60 \text{ seconds} \]

Thus, we can calculate:
\[ \omega_f = 420 \text{ rpm} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{420 \times 2\pi}{60} = 14\pi \text{ rad/s} \]

### Step 2: Use the kinematic equation for rotational motion to find the angular displacement \( \theta \).

The kinematic equation relating final angular velocity, initial angular velocity, angular acceleration, and angular displacement is:
\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]

Where:
- \( \omega_i = 0 \) (initial angular velocity, since the wheel is stationary),
- \( \alpha = 5.5 \text{ rad/s}^2 \) (angular acceleration),
- \( \theta \) is the angular displacement we need to find.

Substituting the known values:
\[ (14\pi)^2 = 0 + 2 \times 5.5 \times \theta \]
\[ 196\pi^2 = 11\theta \]

### Step 3: Solve for \( \theta \).

Rearranging the equation gives:
\[ \theta = \frac{196\pi^2}{11} \]

### Step 4: Calculate the number of rotations.

To find the number of rotations, we need to convert angular displacement \( \theta \) from radians to revolutions. Since one complete revolution is \( 2\pi \) radians, the number of rotations \( N \) is given by:
\[ N = \frac{\theta}{2\pi} \]
Substituting for \( \theta \):
\[ N = \frac{196\pi^2 / 11}{2\pi} = \frac{196\pi}{22} = \frac{98\pi}{11} \]

### Step 5: Calculate the numerical value of \( N \).

Using \( \pi \approx 3.14 \):
\[ N \approx \frac{98 \times 3.14}{11} \approx \frac{307.72}{11} \approx 27.025 \]

Since the number of rotations must be a whole number, we round it to the nearest whole number:
\[ N \approx 28 \]

### Final Answer:
The number of rotations made by the wheel is approximately **28**.

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