A particle moves in a circle with a uniform speed. When it goes from a point A to a diametrically opposite point B, the momentum of the particle changes by `vecP_(A)-vecP_(B)=2` kg m/s `(hatj)` and the centripetal force acting on it changes by `vecF_(A)-vecF_(B)=8N(hati)` where `hati,hatj` are unit vectors along X and Y axes respectively. The angular velocity of the particle is

Options

A. `"8 rad s"^(-1)`
B. `"4 rad s"^(-1)`
C. `"2 rad s"^(-1)`
D. `"16 rad s"^(-1)`

Answer

To solve the problem step by step, we will analyze the changes in momentum and centripetal force as the particle moves from point A to point B.

### Step 1: Understand the momentum change
The momentum of a particle moving in a circle with uniform speed can be expressed as:
- At point A: \( \vec{P}_A = m \vec{v}_A \)
- At point B: \( \vec{P}_B = m \vec{v}_B \)

Given that the particle moves from point A to the diametrically opposite point B, the velocity vectors will be:
- \( \vec{v}_A = v_0 \hat{j} \) (upward direction)
- \( \vec{v}_B = -v_0 \hat{j} \) (downward direction)

Thus, the momentum at points A and B will be:
- \( \vec{P}_A = m v_0 \hat{j} \)
- \( \vec{P}_B = -m v_0 \hat{j} \)

The change in momentum is:
\[ \vec{P}_A - \vec{P}_B = m v_0 \hat{j} - (-m v_0 \hat{j}) = 2 m v_0 \hat{j} \]
According to the problem, this change is given as \( 2 \hat{j} \) kg m/s. Therefore, we can equate:
\[ 2 m v_0 = 2 \quad \Rightarrow \quad m v_0 = 1 \quad \text{(Equation 1)} \]

### Step 2: Understand the centripetal force change
The centripetal force acting on the particle is given by:
- At point A: \( \vec{F}_A = \frac{m v_0^2}{r} (-\hat{i}) \)
- At point B: \( \vec{F}_B = \frac{m v_0^2}{r} \hat{i} \)

The change in centripetal force is:
\[ \vec{F}_A - \vec{F}_B = \left(-\frac{m v_0^2}{r} \hat{i}\right) - \left(\frac{m v_0^2}{r} \hat{i}\right) = -\frac{2 m v_0^2}{r} \hat{i} \]
According to the problem, this change is given as \( 8 \hat{i} \) N. Therefore, we can equate:
\[ -\frac{2 m v_0^2}{r} = 8 \quad \Rightarrow \quad \frac{2 m v_0^2}{r} = -8 \quad \Rightarrow \quad 2 m v_0^2 = 8r \quad \Rightarrow \quad m v_0^2 = 4r \quad \text{(Equation 2)} \]

### Step 3: Relate the two equations
Now we have two equations:
1. \( m v_0 = 1 \) (Equation 1)
2. \( m v_0^2 = 4r \) (Equation 2)

From Equation 1, we can express \( m \) as:
\[ m = \frac{1}{v_0} \]

Substituting this into Equation 2:
\[ \frac{1}{v_0} v_0^2 = 4r \quad \Rightarrow \quad v_0 = 4r \]

### Step 4: Find the angular velocity
The angular velocity \( \omega \) is related to the linear speed \( v_0 \) by:
\[ \omega = \frac{v_0}{r} \]
Substituting \( v_0 = 4r \):
\[ \omega = \frac{4r}{r} = 4 \text{ rad/s} \]

### Final Answer
The angular velocity of the particle is \( \omega = 4 \text{ rad/s} \).

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