A solid body rotates about a stationary axis, so that its angular velocity depends on the rotational angle `phi` as `omega=omega_(0)-kphi` where `omega_(0)` and K are positive constants. At the moment `t=0,phi=0` Find the dependence of rotation angle.

Options

A. `komega_(0)e^(-kt)`
B. `(omega_(0))/(k)e^(-kt)`
C. `(omega_(0))/(k)(1-e^(-kt))`
D. `(k)/(omega_(0))(e^(-kt)-1)`

Answer

To find the dependence of the rotational angle \( \phi \) on time \( t \) for the given angular velocity equation \( \omega = \omega_0 - k\phi \), we can follow these steps:

### Step 1: Relate Angular Velocity to Angular Displacement
We know that angular velocity \( \omega \) is defined as the rate of change of angular displacement \( \phi \) with respect to time \( t \):
\[ \omega = \frac{d\phi}{dt} \]
Substituting the given expression for \( \omega \):
\[ \frac{d\phi}{dt} = \omega_0 - k\phi \]

### Step 2: Rearrange the Equation
Rearranging the equation gives:
\[ dt = \frac{d\phi}{\omega_0 - k\phi} \]

### Step 3: Integrate Both Sides
We will integrate both sides. The left side integrates with respect to time \( t \) from \( 0 \) to \( t \), and the right side integrates with respect to \( \phi \) from \( 0 \) to \( \phi \):
\[ \int_0^t dt = \int_0^{\phi} \frac{d\phi}{\omega_0 - k\phi} \]
This simplifies to:
\[ t = \int_0^{\phi} \frac{d\phi}{\omega_0 - k\phi} \]

### Step 4: Solve the Right Side Integral
The integral on the right side can be solved using the natural logarithm:
\[ \int \frac{d\phi}{\omega_0 - k\phi} = -\frac{1}{k} \ln|\omega_0 - k\phi| \]
Thus, we have:
\[ t = -\frac{1}{k} \left[ \ln|\omega_0 - k\phi| \right]_0^{\phi} \]
This results in:
\[ t = -\frac{1}{k} \left( \ln|\omega_0 - k\phi| - \ln|\omega_0| \right) \]
\[ t = -\frac{1}{k} \ln\left( \frac{\omega_0 - k\phi}{\omega_0} \right) \]

### Step 5: Simplify the Expression
We can rewrite this as:
\[ -\frac{kt}{1} = \ln\left( \frac{\omega_0 - k\phi}{\omega_0} \right) \]
Exponentiating both sides gives:
\[ \frac{\omega_0 - k\phi}{\omega_0} = e^{-kt} \]
Thus:
\[ \omega_0 - k\phi = \omega_0 e^{-kt} \]

### Step 6: Solve for \( \phi \)
Rearranging gives:
\[ k\phi = \omega_0 - \omega_0 e^{-kt} \]
\[ \phi = \frac{\omega_0}{k} (1 - e^{-kt}) \]

### Final Result
The dependence of the rotational angle \( \phi \) on time \( t \) is:
\[ \phi(t) = \frac{\omega_0}{k} (1 - e^{-kt}) \]