A particle of mass m is moving along a trajectory given by x = x0 + a cos(w1 t) and y = y0 + b sin(w

Q: Why do we need acceleration to find torque?

Torque is defined as $\vec{r} \times \vec{F}$. We aren't given the force directly, but we are given the mass and the trajectory. By Newton's second law, $\vec{F} = m\vec{a}$. So, finding the acceleration allows us to find the force, and consequently, the torque.

Q: What happens to the $\hat{i} \times \hat{i}$ term during the cross product?

The cross product of any vector with itself (or a parallel vector) is zero. This is because the magnitude of a cross product depends on the sine of the angle between the vectors. The angle between $\hat{i}$ and $\hat{i}$ is $0^\circ$, and $\sin(0^\circ) = 0$.

Q: Why is $\hat{j} \times \hat{i} = -\hat{k}$?

This comes from the right-hand rule for cross products. If you point the fingers of your right hand in the direction of the positive y-axis ($\hat{j}$) and curl them towards the positive x-axis ($\hat{i}$), your thumb points into the page, which is the negative z-direction ($-\hat{k}$).

Q: What if we were asked for torque at a different time, say $t = \pi / 2\omega_1$?

The process would be exactly the same! You would substitute that specific time into the $\vec{r}$ and $\vec{a}$ equations before taking the cross product. The math would be a bit longer because the sine and cosine terms might not simplify to 0 or 1, but the underlying physics concept is identical.

Q: Is the mass $m$ important here? Can we just find $\vec{r} \times \vec{a}$?

Yes, the mass is crucial. Torque is a dynamic quantity that depends on the actual force applied. $\vec{r} \times \vec{a}$ gives you the 'kinematic' angular acceleration cross product, but multiplying by mass converts it into the physical torque required to cause that motion.

Question

A particle of mass $m$ is moving along a trajectory given by $x = x_0 + a \cos\omega_1 t$ and $y = y_0 + b \sin\omega_2 t$. The torque acting on the particle about the origin at $t = 0$ is

Options:

(A) zero

(B) $m(-x_0 b + y_0 a)\omega_1^2 \hat{k}$

(C) $-m(x_0 b \omega_2^2 - y_0 a \omega_1^2)\hat{k}$

(D) $+my_0 a \omega_1^2 \hat{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle tracing a complex path in the $xy$-plane. We are given its coordinates as functions of time:

$$x = x_0 + a \cos\omega_1 t$$

$$y = y_0 + b \sin\omega_2 t$$

Our goal is to find the torque acting on this particle about the origin at the exact moment $t = 0$. To do this, we need to recall the fundamental definition of torque. Torque $\vec{	au}$ is the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.

By Newton's Second Law, force is mass times acceleration ($\vec{F} = m\vec{a}$). Therefore, we can rewrite the torque equation as:

$$\vec{	au} = m(\vec{r} 	imes \vec{a})$$

This means our mission is clear: we need to find the position vector $\vec{r}$ and the acceleration vector $\vec{a}$ at $t = 0$, and then compute their cross product.

### The Kinematic Journey

To find acceleration, we must differentiate our position equations twice with respect to time. Let's start with velocity. Remember the chain rule!

$$v_x = \frac{dx}{dt} = -a\omega_1 \sin\omega_1 t$$

$$v_y = \frac{dy}{dt} = b\omega_2 \cos\omega_2 t$$

Differentiating once more gives us the acceleration components:

$$a_x = \frac{dv_x}{dt} = -a\omega_1^2 \cos\omega_1 t$$

$$a_y = \frac{dv_y}{dt} = -b\omega_2^2 \sin\omega_2 t$$

### Evaluating at $t = 0$

The problem specifically asks for the torque at $t = 0$. This simplifies our expressions beautifully because $\cos(0) = 1$ and $\sin(0) = 0$. Let's plug $t = 0$ into our position and acceleration equations.

For position:

$$x = x_0 + a(1) = x_0 + a$$

$$y = y_0 + b(0) = y_0$$

So, our position vector is $\vec{r} = (x_0 + a)\hat{i} + y_0\hat{j}$.

For acceleration:

$$a_x = -a\omega_1^2(1) = -a\omega_1^2$$

$$a_y = -b\omega_2^2(0) = 0$$

So, our acceleration vector is $\vec{a} = -a\omega_1^2\hat{i}$.

### The Final Cross Product

Now, we substitute these vectors back into our torque equation:

$$\vec{	au} = m [((x_0 + a)\hat{i} + y_0\hat{j}) 	imes (-a\omega_1^2\hat{i})]$$

When distributing the cross product, remember the rules for unit vectors: $\hat{i} 	imes \hat{i} = 0$ (a vector crossed with itself is zero), and $\hat{j} 	imes \hat{i} = -\hat{k}$ (by the right-hand rule).

The $\hat{i} 	imes \hat{i}$ term vanishes, leaving us with:

$$\vec{	au} = m [y_0 (-a\omega_1^2) (\hat{j} 	imes \hat{i})]$$

$$\vec{	au} = m (-y_0 a \omega_1^2) (-\hat{k})$$

The two negative signs cancel out, yielding our final, elegant result:

**$$\vec{	au} = +my_0 a \omega_1^2 \hat{k}$$**

Animated Solution for Physics - Rotational Motion: A particle of mass $m$ is moving along a trajectory given by $x = x_{0} + a cos ω_{1} t$ and $y = y_{0} + b sin ω_{2} t$ . The torque acting on the particle about the origin at $t = 0$ is

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Visualized Solution

\text{Trajectory Equations}

\text{Torque Formula}

\text{Velocity Components}

\text{Acceleration Components}

\text{Vectors at } t = 0

\text{Calculating Torque}

\text{Final Torque}

The Sigma Insight: Torque and Angular Momentum

Analyzing the Setup

The Kinematic Journey

Evaluating at $t = 0$

The Final Cross Product

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