The coordinates of a particle of mass `'m'` as function of time are given by `x=x_(0)+a_(1) cos(omegat)` and `y=y_(0)+a_(2)sin(omega_(2)t)`. The torque on particle about origin at time `t=0` is :

Options

A. `(ma_(1)omega_(1)^(2)x_(0))k^(hat)`
B. `(ma_(1)omega_(1)^(2)x_(0)^(2))k`
C. `(ma_(1)omega_(1)^(2)y_(0))k`
D. `(ma_(1)omega_(1)^(2)x_(0)y_(0))k`

Answer

To solve the problem, we need to find the torque on a particle about the origin at time \( t = 0 \). The coordinates of the particle are given as:

\[ x = x_0 + a_1 \cos(\omega t) \]
\[ y = y_0 + a_2 \sin(\omega_2 t) \]

### Step 1: Determine the position vector \( \mathbf{R} \) at \( t = 0 \)

At \( t = 0 \):

\[ x(0) = x_0 + a_1 \cos(0) = x_0 + a_1 \]
\[ y(0) = y_0 + a_2 \sin(0) = y_0 \]

Thus, the position vector \( \mathbf{R} \) at \( t = 0 \) is:

\[ \mathbf{R} = (x_0 + a_1) \hat{i} + y_0 \hat{j} \]

### Step 2: Find the velocity vector \( \mathbf{V} \)

To find the force, we first need the acceleration. We start by finding the velocity vector \( \mathbf{V} \) by differentiating the position vector with respect to time:

\[ \frac{dx}{dt} = -a_1 \omega \sin(\omega t) \]
\[ \frac{dy}{dt} = a_2 \omega_2 \cos(\omega_2 t) \]

Thus, the velocity vector \( \mathbf{V} \) at \( t = 0 \) is:

\[ \mathbf{V}(0) = -a_1 \omega \sin(0) \hat{i} + a_2 \omega_2 \cos(0) \hat{j} = 0 \hat{i} + a_2 \omega_2 \hat{j} = a_2 \omega_2 \hat{j} \]

### Step 3: Find the acceleration vector \( \mathbf{A} \)

Now, we differentiate the velocity vector to find the acceleration vector \( \mathbf{A} \):

\[ \frac{d^2x}{dt^2} = -a_1 \omega^2 \cos(\omega t) \]
\[ \frac{d^2y}{dt^2} = -a_2 \omega_2^2 \sin(\omega_2 t) \]

At \( t = 0 \):

\[ \mathbf{A}(0) = -a_1 \omega^2 \cos(0) \hat{i} - a_2 \omega_2^2 \sin(0) \hat{j} = -a_1 \omega^2 \hat{i} + 0 \hat{j} = -a_1 \omega^2 \hat{i} \]

### Step 4: Find the force vector \( \mathbf{F} \)

Using Newton's second law, \( \mathbf{F} = m \mathbf{A} \):

\[ \mathbf{F} = m (-a_1 \omega^2 \hat{i}) = -m a_1 \omega^2 \hat{i} \]

### Step 5: Calculate the torque \( \mathbf{\tau} \)

The torque \( \mathbf{\tau} \) about the origin is given by:

\[ \mathbf{\tau} = \mathbf{R} \times \mathbf{F} \]

Substituting the values of \( \mathbf{R} \) and \( \mathbf{F} \):

\[ \mathbf{R} = (x_0 + a_1) \hat{i} + y_0 \hat{j} \]
\[ \mathbf{F} = -m a_1 \omega^2 \hat{i} \]

Now, we compute the cross product:

\[ \mathbf{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_0 + a_1 & y_0 & 0 \\ -m a_1 \omega^2 & 0 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{\tau} = \hat{k} \left[ (x_0 + a_1)(0) - (y_0)(-m a_1 \omega^2) \right] = m a_1 \omega^2 y_0 \hat{k} \]

### Final Answer

Thus, the torque on the particle about the origin at time \( t = 0 \) is:

\[ \mathbf{\tau} = m a_1 \omega^2 y_0 \hat{k} \]