A particle of mass m is moving along a trajectory given by
`x=x_0+a cosomega_1t`
`y=y_0+bsinomega_2t`
The torque, acting on the particle about the origin, at `t=0` is:

Options

A. a) `m(-x_0b+y_0a)omega_1^2hatk`
B. b) `+my_0aomega_1^2hatk`
C. c) Zero
D. d) `-m(x_0bomega_2^2-y_0aomega_1^2)hatk`

Answer

To find the torque acting on a particle of mass \( m \) moving along the given trajectory at \( t = 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position Vector**:
The position vector \( \mathbf{r} \) of the particle is given by:
\[ \mathbf{r} = x \hat{i} + y \hat{j} \]
where
\[ x = x_0 + a \cos(\omega_1 t) \quad \text{and} \quad y = y_0 + b \sin(\omega_2 t) \]

2. **Calculate the Acceleration**:
The acceleration \( \mathbf{a} \) can be found by taking the second derivative of the position vector \( \mathbf{r} \) with respect to time \( t \):
\[ \mathbf{a} = \frac{d^2 \mathbf{r}}{dt^2} \]
First, we find the velocity \( \mathbf{v} \):
\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} = \left(-a \omega_1 \sin(\omega_1 t) \hat{i} + b \omega_2 \cos(\omega_2 t) \hat{j}\right) \]
Now, taking the derivative again to find acceleration:
\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} = \left(-a \omega_1^2 \cos(\omega_1 t) \hat{i} - b \omega_2^2 \sin(\omega_2 t) \hat{j}\right) \]

3. **Calculate the Force**:
The force \( \mathbf{F} \) acting on the particle is given by Newton's second law:
\[ \mathbf{F} = m \mathbf{a} = m \left(-a \omega_1^2 \cos(\omega_1 t) \hat{i} - b \omega_2^2 \sin(\omega_2 t) \hat{j}\right) \]

4. **Calculate the Torque**:
The torque \( \mathbf{\tau} \) about the origin is given by:
\[ \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \]
At \( t = 0 \):
\[ x(0) = x_0 + a \cos(0) = x_0 + a \]
\[ y(0) = y_0 + b \sin(0) = y_0 \]
Therefore, the position vector at \( t = 0 \) is:
\[ \mathbf{r}(0) = (x_0 + a) \hat{i} + y_0 \hat{j} \]
The force at \( t = 0 \) is:
\[ \mathbf{F}(0) = m \left(-a \omega_1^2 \cos(0) \hat{i} - b \omega_2^2 \sin(0) \hat{j}\right) = -m a \omega_1^2 \hat{i} \]
Now, substituting into the torque equation:
\[ \mathbf{\tau} = \mathbf{r}(0) \times \mathbf{F}(0) \]
\[ \mathbf{\tau} = \left((x_0 + a) \hat{i} + y_0 \hat{j}\right) \times \left(-m a \omega_1^2 \hat{i}\right) \]
The cross product results in:
\[ \mathbf{\tau} = (x_0 + a)(-m a \omega_1^2) \hat{k} - y_0(0) = -m a \omega_1^2 (x_0 + a) \hat{k} \]

5. **Final Result**:
Thus, the torque acting on the particle about the origin at \( t = 0 \) is:
\[ \mathbf{\tau} = -m a \omega_1^2 (x_0 + a) \hat{k} \]