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Step-by-Step Solution
Step 1: Write down the equations of motion
The particle's position at time t is given by:
$x = x_0 + a \cos(\omega_1 t)$
$y = y_0 + b \sin(\omega_2 t)$
Step 2: Find the acceleration at t = 0
(a) Differentiate $x(t)$ and $y(t)$ twice to get accelerations:
$ \frac{d^2 x}{dt^2} = -\,a\,\omega_1^2 \cos(\omega_1 t), \; \frac{d^2 y}{dt^2} = -\,b\,\omega_2^2 \sin(\omega_2 t).$
(b) At $t = 0$,
$ \frac{d^2 x}{dt^2}\Big|_{t=0} = -\,a\,\omega_1^2, \quad \frac{d^2 y}{dt^2}\Big|_{t=0} = 0 \quad(\text{since } \sin(0) = 0). $
Hence the acceleration vector at $t=0$ is:
$ \overrightarrow{a}(0) = -\,a\,\omega_1^2\,\hat{i} + 0\,\hat{j}. $
Therefore the force at $t=0$ is:
$ \overrightarrow{F}(0) = m \,\overrightarrow{a}(0) = -\,m\,a\,\omega_1^2\,\hat{i}. $
Step 3: Find the position vector at t = 0
Substitute $t=0$ into $x(t)$ and $y(t)$:
$ x(0) = x_0 + a \,\cos(0) = x_0 + a, \quad y(0) = y_0 + b \,\sin(0) = y_0. $
Hence the position vector at $t=0$ is:
$ \overrightarrow{r}(0) = (x_0 + a)\,\hat{i} + y_0\,\hat{j}. $
Step 4: Compute the torque about the origin at t = 0
The torque is given by $ \overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}. $
At $ t=0 $:
$ \overrightarrow{\tau}(0)
= \big[(x_0 + a)\,\hat{i} + y_0\,\hat{j}\big]
\times \big[-\,m\,a\,\omega_1^2\,\hat{i}\big]. $
Using the cross product:
$
\overrightarrow{r}(0) \times \overrightarrow{F}(0)
= \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
x_0 + a & y_0 & 0 \\
-m\,a\,\omega_1^2 & 0 & 0
\end{vmatrix}.
$
The determinant yields a component only in the $ \hat{k} $ direction:
$ \tau_z = (x_0 + a)\times 0 \;-\; y_0 \big(-\,m\,a\,\omega_1^2 \big)
= m\,y_0\,a\,\omega_1^2. $
Therefore,
$ \overrightarrow{\tau}(0) = m\,y_0\,a\,\omega_1^2 \,\hat{k}. $
Step 5: State the final answer
The torque acting on the particle about the origin at $t=0$ is:
$ +\,m\,y_0\,a\,\omega_1^2 \,\hat{k}. $
