Let's solve the problem step by step.
### Given Data:
- Radius of the circle, \( r = 1 \, \text{m} \)
- Initial angular speed, \( \omega_0 = 12 \, \text{rad/s} \)
- Final angular speed after 2 seconds, \( \omega = 480\pi \, \text{rpm} \)
- Time interval, \( t = 2 \, \text{s} \)
### Step 1: Convert Final Angular Speed to Radians per Second
To convert from rpm to rad/s, we use the conversion factor:
\[
\omega = 480\pi \, \text{rpm} = 480\pi \times \frac{2\pi}{60} \, \text{rad/s} = 16 \, \text{rad/s}
\]
### Step 2: Calculate Angular Acceleration
Using the equation of motion for angular velocity:
\[
\omega = \omega_0 + \alpha t
\]
Substituting the known values:
\[
16 = 12 + \alpha \cdot 2
\]
Rearranging gives:
\[
\alpha \cdot 2 = 16 - 12
\]
\[
\alpha \cdot 2 = 4 \implies \alpha = 2 \, \text{rad/s}^2
\]
### Step 3: Tangential Velocity as a Function of Time
The angular velocity as a function of time is given by:
\[
\omega(t) = \omega_0 + \alpha t = 12 + 2t
\]
The tangential velocity \( v \) is related to angular velocity by:
\[
v(t) = \omega(t) \cdot r = (12 + 2t) \cdot 1 = 12 + 2t
\]
This is valid for \( t \leq 2 \). For \( t > 2 \):
\[
v(t) = \omega(2) \cdot r = 16 \cdot 1 = 16 \, \text{m/s}
\]
### Step 4: Acceleration of the Particle at \( t = 0.5 \, \text{s} \) and \( t = 3 \, \text{s} \)
The total acceleration \( a \) has two components: tangential acceleration \( a_t \) and normal acceleration \( a_n \).
1. **At \( t = 0.5 \, \text{s} \)**:
- Calculate \( \omega \) at \( t = 0.5 \):
\[
\omega(0.5) = 12 + 2 \cdot 0.5 = 13 \, \text{rad/s}
\]
- Tangential acceleration:
\[
a_t = \alpha \cdot r = 2 \cdot 1 = 2 \, \text{m/s}^2
\]
- Normal acceleration:
\[
a_n = \omega^2 r = 13^2 \cdot 1 = 169 \, \text{m/s}^2
\]
- Total acceleration:
\[
a = \sqrt{a_t^2 + a_n^2} = \sqrt{2^2 + 169^2} = \sqrt{4 + 28561} = \sqrt{28565} \approx 169.01 \, \text{m/s}^2
\]
2. **At \( t = 3 \, \text{s} \)**:
- After \( t = 2 \, \text{s} \), the angular acceleration is zero, so:
\[
a_t = 0
\]
- The angular velocity remains constant:
\[
\omega(3) = 16 \, \text{rad/s}
\]
- Normal acceleration:
\[
a_n = \omega^2 r = 16^2 \cdot 1 = 256 \, \text{m/s}^2
\]
- Total acceleration:
\[
a = \sqrt{0^2 + 256^2} = 256 \, \text{m/s}^2
\]
### Step 5: Angular Displacement at \( t = 3 \, \text{s} \)
The angular displacement \( \theta \) can be calculated in two parts:
1. From \( t = 0 \) to \( t = 2 \):
\[
\theta_1 = \omega_0 t + \frac{1}{2} \alpha t^2 = 12 \cdot 2 + \frac{1}{2} \cdot 2 \cdot 2^2 = 24 + 4 = 28 \, \text{radians}
\]
2. From \( t = 2 \) to \( t = 3 \):
\[
\theta_2 = \omega(2) \cdot (3 - 2) = 16 \cdot 1 = 16 \, \text{radians}
\]
3. Total angular displacement:
\[
\theta = \theta_1 + \theta_2 = 28 + 16 = 44 \, \text{radians}
\]
### Summary of Results:
(i) Angular acceleration, \( \alpha = 2 \, \text{rad/s}^2 \)
(ii) Tangential velocity as a function of time:
\[
v(t) = 12 + 2t \quad \text{for } t \leq 2 \quad \text{and } v(t) = 16 \, \text{m/s} \quad \text{for } t > 2
\]
(iii) Acceleration at \( t = 0.5 \, \text{s} \): \( a \approx 169.01 \, \text{m/s}^2 \) and at \( t = 3 \, \text{s} \): \( a = 256 \, \text{m/s}^2 \)
(iv) Angular displacement at \( t = 3 \, \text{s} \): \( \theta = 44 \, \text{radians} \)
