Animated Solution for Physics - Rotational Motion: The position vector $\mathbf{r}$ of particle of mass $m$ is given by the following equation $\mathbf{r}(t)=\alpha t^3\hat{\mathbf{i}}+\beta t^2\hat{\mathbf{j}}$ where, $\alpha =\frac{10}{3}{\text{ ms}}^{-3}$, $\beta =5{\text{ ms}}^{-2}$ and $m=0.1\text{ kg}$. At $t=1\text{s}$, which of the following statement(s) is (are) true about the particle?

The journey of a particle through space is often described by a simple mathematical construct: the position vector. In this problem, we are given the position vector of a particle as a function of time, and we need to unravel its entire dynamic profile—its velocity, acceleration, force, angular momentum, and torque.

This is a classic exercise in kinematics and dynamics, where calculus serves as our primary tool to bridge the gap between position and motion.

Imagine the particle moving along a curved path in the $x-y$ plane. We are given its position vector:

Here, the constants are $\alpha =\frac{10}{3}{\text{ ms}}^{-3}$ and $\beta =5{\text{ ms}}^{-2}$. The mass of the particle is $m=0.1\text{ kg}$.

Our goal is to evaluate various physical quantities at the specific instant $t=1\text{s}$. At this moment, the particle's position is simply:

This vector points from the origin to the particle's location in the $x-y$ plane.

To find the velocity, force, angular momentum, and torque, we must rely on the fundamental relationships of classical mechanics.

Velocity is the rate of change of position, so we differentiate $\mathbf{r}(t)$ with respect to time.

Acceleration is the rate of change of velocity, requiring a second differentiation.

Once we have acceleration, Newton's Second Law ($\mathbf{F}=m\mathbf{a}$) gives us the force.

For rotational dynamics, Angular Momentum is defined as the cross product of the position vector and linear momentum ($\mathbf{L}=\mathbf{r}\times \mathbf{p}$), and Torque is the cross product of the position vector and force ($\tau =\mathbf{r}\times \mathbf{F}$).

Let's execute the first derivative to find the velocity vector:

Using the power rule, the derivative of $t^3$ is $3t^2$, and the derivative of $t^2$ is $2t$. This yields:

Now, we substitute $t=1\text{s}$, along with the given values of $\alpha$ and $\beta$:

Simplifying this expression, we get:

This perfectly matches option (a). The particle is moving diagonally in the positive $x$ and $y$ directions with equal speed components.

Next, we need the force. To get there, we must first find the acceleration by differentiating the velocity vector:

Evaluating this at $t=1\text{s}$:

Now, we apply Newton's Second Law. Multiplying the acceleration by the mass $m=0.1\text{ kg}$:

Looking at option (c), it claims the force is $\hat{\mathbf{i}}+2\hat{\mathbf{j}}$. This is a classic trap—the components are swapped! Therefore, option (c) is incorrect.

Now we step into the rotational realm. The angular momentum $\mathbf{L}$ about the origin is given by:

We already have $\mathbf{r}(1)$ and $\mathbf{v}(1)$. Let's set up the cross product:

When computing the cross product, remember the cyclic rules of unit vectors: $\hat{\mathbf{i}}\times \hat{\mathbf{j}}=\hat{\mathbf{k}}$ and $\hat{\mathbf{j}}\times \hat{\mathbf{i}}=-\hat{\mathbf{k}}$. The cross product of a vector with itself is zero.

Taking the common denominator:

This confirms that option (b) is absolutely correct.

Finally, let's evaluate the torque $\tau$ acting on the particle about the origin:

Using our previously calculated vectors for position and force:

Expanding the cross product:

This matches option (d) perfectly.

We have found that options (a), (b), and (d) are correct. But let's look deeper into the physics.

Notice that both the angular momentum and the torque have a negative $\hat{\mathbf{k}}$ component. This means both vectors point into the page (the $-z$ direction).

Since torque is the time derivative of angular momentum ($\tau =\frac{d\mathbf{L}}{dt}$), the fact that they point in the same direction implies that the magnitude of the angular momentum is increasing over time. The force is actively "twisting" the particle faster around the origin!