A disk in xy plane, initially at rest, of mass 1 kg and radius 10 m, with center at origin, can only rotate about its axis, that is, about z-axis due to a fixed axle through its center. A force `vecF = (3hati -4hatj+hatk)` N starts acting on it at the point `(3hati +5hatj)` m. The work done by the force in the first one second is

Options

A. 2.39 J
B. 3.27 J
C. 9.27 J
D. 7.29 J

Answer

To solve the problem, we need to find the work done by the force on the disk in the first second. Let's break it down step by step.

### Step 1: Identify the Given Data
- Mass of the disk, \( m = 1 \, \text{kg} \)
- Radius of the disk, \( R = 10 \, \text{m} \)
- Force acting on the disk, \( \vec{F} = 3\hat{i} - 4\hat{j} + \hat{k} \, \text{N} \)
- Point of application of the force, \( \vec{r} = 3\hat{i} + 5\hat{j} \, \text{m} \)
- Time, \( t = 1 \, \text{s} \)

### Step 2: Calculate the Torque
Torque (\( \vec{\tau} \)) is given by the cross product of the position vector (\( \vec{r} \)) and the force vector (\( \vec{F} \)):
\[ \vec{\tau} = \vec{r} \times \vec{F} \]

Calculating the cross product:
\[ \vec{r} = (3, 5, 0) \quad \text{and} \quad \vec{F} = (3, -4, 1) \]
Using the determinant to find the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 0 \\ 3 & -4 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{\tau} = \hat{i}(5 \cdot 1 - 0 \cdot (-4)) - \hat{j}(3 \cdot 1 - 0 \cdot 3) + \hat{k}(3 \cdot (-4) - 5 \cdot 3) \]
\[ = \hat{i}(5) - \hat{j}(3) + \hat{k}(-12 - 15) \]
\[ = 5\hat{i} - 3\hat{j} - 27\hat{k} \, \text{N m} \]

### Step 3: Calculate the Magnitude of Torque
The magnitude of the torque is:
\[ |\vec{\tau}| = \sqrt{(5)^2 + (-3)^2 + (-27)^2} = \sqrt{25 + 9 + 729} = \sqrt{763} \approx 27.6 \, \text{N m} \]

### Step 4: Calculate Angular Acceleration
Using the relation between torque and angular acceleration (\( \alpha \)):
\[ \tau = I \alpha \]
For a disk, the moment of inertia \( I \) about the z-axis is:
\[ I = \frac{1}{2} m R^2 = \frac{1}{2} \cdot 1 \cdot (10)^2 = 50 \, \text{kg m}^2 \]
So,
\[ \alpha = \frac{\tau}{I} = \frac{|\vec{\tau}|}{I} = \frac{27.6}{50} \approx 0.552 \, \text{rad/s}^2 \]

### Step 5: Calculate Final Angular Velocity
Since the disk starts from rest, the initial angular velocity \( \omega_0 = 0 \). The final angular velocity after 1 second is:
\[ \omega = \omega_0 + \alpha t = 0 + 0.552 \cdot 1 = 0.552 \, \text{rad/s} \]

### Step 6: Calculate Work Done
The work done by the force is equal to the change in rotational kinetic energy:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2} I \omega^2 - 0 \]
Calculating the final kinetic energy:
\[ KE_{\text{final}} = \frac{1}{2} \cdot 50 \cdot (0.552)^2 \approx \frac{1}{2} \cdot 50 \cdot 0.304 \approx 7.6 \, \text{J} \]

Thus, the work done by the force in the first second is approximately:
\[ \boxed{7.6 \, \text{J}} \]