A particle describes uniform circular motion in a circle of radius 2 m, with the angular speed of 2 rad `s^(-1)`. The magnitude of the change in its velocity in `pi/2` s is

Options

A. `0 m s^(-1)`
B. `2sqrt2ms^(-1)`
C. `8 m s^(-1)`
D. `4 m s^(-1)`

Answer

To solve the problem step by step, we will follow the concepts of circular motion and the relationships between angular and linear quantities.

### Step 1: Determine the angular displacement (θ)
Given:
- Angular speed (ω) = 2 rad/s
- Time (t) = π/2 s

Using the formula for angular displacement:
\[ \theta = \omega \times t \]
Substituting the values:
\[ \theta = 2 \, \text{rad/s} \times \frac{\pi}{2} \, \text{s} = \pi \, \text{radians} \]

### Step 2: Calculate the linear velocity (v)
The linear velocity (v) of a particle in circular motion is given by:
\[ v = r \times \omega \]
Where:
- Radius (r) = 2 m
- Angular speed (ω) = 2 rad/s

Substituting the values:
\[ v = 2 \, \text{m} \times 2 \, \text{rad/s} = 4 \, \text{m/s} \]

### Step 3: Calculate the change in velocity (Δv)
The change in velocity for a particle moving in a circular path can be calculated using the formula:
\[ \Delta v = 2v \times \sin\left(\frac{\theta}{2}\right) \]
Substituting the values:
- We already calculated \(v = 4 \, \text{m/s}\)
- \(\theta = \pi \, \text{radians}\)

Now, calculate \(\sin\left(\frac{\theta}{2}\right)\):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]

Now substituting into the change in velocity formula:
\[ \Delta v = 2 \times 4 \, \text{m/s} \times 1 = 8 \, \text{m/s} \]

### Final Answer:
The magnitude of the change in velocity in \(\frac{\pi}{2}\) seconds is **8 m/s**.

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