Let's solve the question step by step.
### Given Data:
- Radius of the circle, \( R = 1 \, \text{m} \)
- Initial angular speed, \( \omega_i = 12 \, \text{rad/s} \)
- Final angular speed in rpm, \( n = \frac{480}{\pi} \, \text{rpm} \)
- Time duration, \( t = 2 \, \text{s} \)
### Step 1: Convert final angular speed to rad/s
To convert the final angular speed from rpm to rad/s, we use the formula:
\[
\omega_f = 2\pi \cdot \frac{n}{60}
\]
Substituting the value of \( n \):
\[
\omega_f = 2\pi \cdot \frac{480/\pi}{60} = 16 \, \text{rad/s}
\]
### Step 2: Calculate angular acceleration (\( \alpha \))
Using the first kinematic equation for angular motion:
\[
\omega_f = \omega_i + \alpha t
\]
Rearranging to find \( \alpha \):
\[
\alpha = \frac{\omega_f - \omega_i}{t}
\]
Substituting the known values:
\[
\alpha = \frac{16 - 12}{2} = \frac{4}{2} = 2 \, \text{rad/s}^2
\]
### Step 3: Tangential velocity as a function of time
The tangential velocity \( v(t) \) can be expressed as:
\[
v(t) = R \cdot \omega(t)
\]
Where \( \omega(t) = \omega_i + \alpha t \):
\[
\omega(t) = 12 + 2t
\]
Thus,
\[
v(t) = R \cdot (12 + 2t) = 1 \cdot (12 + 2t) = 12 + 2t \, \text{m/s}
\]
### Step 4: Calculate acceleration at \( t = 0.5 \, \text{s} \) and \( t = 3 \, \text{s} \)
The total acceleration \( a \) is given by:
\[
a = \sqrt{a_t^2 + a_r^2}
\]
Where:
- Tangential acceleration \( a_t = R \alpha \)
- Radial acceleration \( a_r = R \omega^2 \)
#### At \( t = 0.5 \, \text{s} \):
1. Calculate \( \omega(0.5) \):
\[
\omega(0.5) = 12 + 2(0.5) = 12 + 1 = 13 \, \text{rad/s}
\]
2. Calculate \( a_t \):
\[
a_t = R \alpha = 1 \cdot 2 = 2 \, \text{m/s}^2
\]
3. Calculate \( a_r \):
\[
a_r = R \omega(0.5)^2 = 1 \cdot (13)^2 = 169 \, \text{m/s}^2
\]
4. Calculate total acceleration:
\[
a = \sqrt{(2)^2 + (169)^2} = \sqrt{4 + 28561} = \sqrt{28565} \approx 169.0 \, \text{m/s}^2
\]
#### At \( t = 3 \, \text{s} \):
After \( t = 2 \, \text{s} \), the particle moves with constant speed \( \omega_f = 16 \, \text{rad/s} \):
1. Calculate \( a_r \):
\[
a_r = R \omega_f^2 = 1 \cdot (16)^2 = 256 \, \text{m/s}^2
\]
2. Since \( a_t = 0 \) (constant speed):
\[
a = a_r = 256 \, \text{m/s}^2
\]
### Step 5: Calculate angular displacement at \( t = 3 \, \text{s} \)
Using the second kinematic equation:
\[
\theta = \omega_i t + \frac{1}{2} \alpha t^2
\]
Substituting the values:
\[
\theta = 12 \cdot 3 + \frac{1}{2} \cdot 2 \cdot (3)^2 = 36 + \frac{1}{2} \cdot 2 \cdot 9 = 36 + 9 = 45 \, \text{radians}
\]
### Summary of Results:
1. Angular acceleration \( \alpha = 2 \, \text{rad/s}^2 \)
2. Tangential velocity \( v(t) = 12 + 2t \, \text{m/s} \)
3. Acceleration at \( t = 0.5 \, \text{s} \) is approximately \( 169.0 \, \text{m/s}^2 \) and at \( t = 3 \, \text{s} \) is \( 256 \, \text{m/s}^2 \)
4. Angular displacement at \( t = 3 \, \text{s} \) is \( 45 \, \text{radians} \)
