A whel of mass 2kg an radius 10 cm is rotating about its axis at an angular velocity of `2pirad//s` the force that must be applied tangentially to the wheel to stop it in 5 revolutions is

Options

A. `3pixx10^(-2)kg-m^(2)//s`
B. `4pixx10^(-2)kg-m^(2)//s`
C. `pixx10^(-2)kg-m^(2)//s`
D. `2pixx10^(-2)kg-m^(2)//s`

Explanation

To solve the problem of finding the tangential force required to stop a rotating wheel in a given number of revolutions, we can follow these steps:

### Step 1: Understand the given data
- Mass of the wheel (m) = 2 kg
- Radius of the wheel (r) = 10 cm = 0.1 m
- Initial angular velocity (ω₀) = 2π rad/s
- Number of revolutions (n) = 5

### Step 2: Convert revolutions to radians
To find the total angle (θ) in radians through which the wheel rotates, we use the formula:
\[ \theta = n \times 2\pi \]
Substituting the values:
\[ \theta = 5 \times 2\pi = 10\pi \text{ radians} \]

### Step 3: Use the rotational kinematics equation
We need to find the angular deceleration (α) required to stop the wheel. We can use the equation of motion for rotational motion:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Since the final angular velocity (ω) is 0 (the wheel stops), we can rearrange the equation to find α:
\[ 0 = (2\pi)^2 + 2\alpha(10\pi) \]
This simplifies to:
\[ 0 = 4\pi^2 + 20\pi\alpha \]
Solving for α:
\[ 20\pi\alpha = -4\pi^2 \implies \alpha = -\frac{4\pi^2}{20\pi} = -\frac{4\pi}{20} = -\frac{\pi}{5} \text{ rad/s}^2 \]

### Step 4: Calculate the moment of inertia (I)
For a solid disk (which we assume the wheel to be), the moment of inertia is given by:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 2 \times (0.1)^2 = \frac{1}{2} \times 2 \times 0.01 = 0.01 \text{ kg m}^2 \]

### Step 5: Relate torque (τ) to force (F)
The torque is related to the force applied tangentially at the radius:
\[ \tau = F \cdot r \]
And torque is also related to moment of inertia and angular acceleration:
\[ \tau = I \cdot \alpha \]
Setting these equal gives:
\[ F \cdot r = I \cdot \alpha \]
Substituting the known values:
\[ F \cdot 0.1 = 0.01 \cdot \left(-\frac{\pi}{5}\right) \]
Solving for F:
\[ F = \frac{0.01 \cdot \left(-\frac{\pi}{5}\right)}{0.1} = -\frac{0.01\pi}{0.5} = -0.02\pi \text{ N} \]

### Step 6: Final answer
The force that must be applied tangentially to the wheel to stop it in 5 revolutions is:
\[ F = 0.02\pi \text{ N} \quad (\text{approximately } 0.0628 \text{ N}) \]