A particle of mass m is moving in time t on a trajectory given by r = 10 alpha t^2 i + 5 beta (t - 5

Q: Why is the initial angular momentum zero?

At $t=0$, both the position vector and the velocity vector lie entirely along the y-axis. Since they are parallel, the angle between them is zero, and their cross product vanishes.

Q: Does the mass of the particle affect the final answer?

No. The mass $m$ is a constant scalar multiplier in the angular momentum equation $\mathbf{L} = m(\mathbf{r} \times \mathbf{v})$. When we set $\mathbf{L}(t) = 0$, the mass cancels out.

Q: What does it physically mean when $\mathbf{r} \times \mathbf{v} = 0$?

It means the velocity vector is purely radial. The particle is moving directly towards or away from the origin, with no tangential component to create rotation around the origin.

Q: Why did we ignore $t=0$ when solving the final equation?

The equation $t^2 = 2t(t-5)$ has $t=0$ as a trivial solution, which corresponds to our initial state. We are looking for the *next* instance in time, so we assume $t > 0$ and divide by $t$.

Q: Can I use the determinant method for 2D vectors?

Absolutely! You can treat 2D vectors as 3D vectors with a zero z-component. The cross product will then purely be in the $\hat{\mathbf{k}}$ direction, making it easy to solve.

Question

A particle of mass $m$ is moving in time $t$ on a trajectory given by
$\mathbf{r} = 10\alpha t^2 \hat{\mathbf{i}} + 5\beta(t - 5)\hat{\mathbf{j}}$
where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for $t = 0$ at time $t$ is ...... s.

Sigma Percentile · Accepted Answer

### The Physical Setup: A Particle in Motion

Imagine a particle of mass $m$ tracing a curved path in a two-dimensional plane. The position of this particle at any given time $t$ is dictated by the vector equation:

$$\mathbf{r}(t) = 10\alpha t^2 \hat{\mathbf{i}} + 5\beta(t - 5)\hat{\mathbf{j}}$$

Here, $\alpha$ and $\beta$ are simply dimensional constants that ensure our units make sense. Our mission is to find the exact moment in time when the particle's angular momentum returns to the exact value it had at the very beginning of its journey, at $t = 0$.

### Unlocking the Angular Momentum

To understand the angular momentum of a particle, we must first know how fast and in what direction it is moving. The angular momentum $\mathbf{L}$ about the origin is defined as the cross product of the position vector $\mathbf{r}$ and the linear momentum $\mathbf{p}$. Since $\mathbf{p} = m\mathbf{v}$, we can write:

$$\mathbf{L} = m(\mathbf{r} 	imes \mathbf{v})$$

To find the velocity vector $\mathbf{v}(t)$, we take the time derivative of the position vector $\mathbf{r}(t)$:

$$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 20\alpha t \hat{\mathbf{i}} + 5\beta \hat{\mathbf{j}}$$

### The Initial State: A Moment of Alignment

Before we can find when the angular momentum returns to its initial value, we must determine what that initial value actually is. Let's evaluate our position and velocity vectors at $t = 0$:

$$\mathbf{r}(0) = 10\alpha(0)^2 \hat{\mathbf{i}} + 5\beta(0 - 5)\hat{\mathbf{j}} = -25\beta \hat{\mathbf{j}}$$

$$\mathbf{v}(0) = 20\alpha(0) \hat{\mathbf{i}} + 5\beta \hat{\mathbf{j}} = 5\beta \hat{\mathbf{j}}$$

Notice something fascinating here? Both the position vector and the velocity vector lie entirely along the y-axis (the $\hat{\mathbf{j}}$ direction). Because they are perfectly parallel, the angle between them is zero. The cross product of two parallel vectors is always zero, which means:

$$\mathbf{L}(0) = m(-25\beta \hat{\mathbf{j}} 	imes 5\beta \hat{\mathbf{j}}) = 0$$

### The Quest for Zero Angular Momentum

The problem asks us to find the time $t$ when $\mathbf{L}(t) = \mathbf{L}(0)$. Since we just discovered that $\mathbf{L}(0) = 0$, we are essentially looking for the time when the angular momentum vanishes again. This happens when the cross product of $\mathbf{r}(t)$ and $\mathbf{v}(t)$ is zero:

$$\mathbf{r}(t) 	imes \mathbf{v}(t) = 0$$

Let's set up the cross product using the determinant method for our 2D vectors:

$$(10\alpha t^2 \hat{\mathbf{i}} + 5\beta(t - 5)\hat{\mathbf{j}}) 	imes (20\alpha t \hat{\mathbf{i}} + 5\beta \hat{\mathbf{j}}) = 0$$

To compute this, we multiply the x-component of $\mathbf{r}$ with the y-component of $\mathbf{v}$, and subtract the product of the y-component of $\mathbf{r}$ and the x-component of $\mathbf{v}$:

$$(10\alpha t^2)(5\beta) - (5\beta(t - 5))(20\alpha t) = 0$$

### The Mathematical Resolution

Now, we embark on a simple algebraic simplification. Expanding the terms, we get:

$$50\alpha\beta t^2 - 100\alpha\beta t(t - 5) = 0$$

Let's move the negative term to the right side of the equation:

$$50\alpha\beta t^2 = 100\alpha\beta t(t - 5)$$

**Don't rush through this!** Notice how the constants $50\alpha\beta$ appear on both sides. We can divide both sides by $50\alpha\beta t$. We can safely divide by $t$ because we are looking for a time $t > 0$ (the *next* time the angular momentum is zero, not the initial time).

$$t = 2(t - 5)$$

Expanding the right side:

$$t = 2t - 10$$

Rearranging the terms yields our final, elegant result:

$$t = 10 	ext{ s}$$

At exactly $t = 10$ seconds, the particle's position and velocity vectors align perfectly once again, bringing the angular momentum back to zero.

Enter Numerical Value:

Visualized Solution

$r (t) = 10 α t^{2} \hat{i} + 5 β (t - 5) \hat{j}$

$L = m (r \times v)$

$v (t) = \frac{d r}{d t}$

$L (0) = 0$

$L (t) = L (0)$

$r \times v = 0$

$Expanding the Cross Product$

$Simplifying the Equation$

$t = 10 s$

The Sigma Insight: Torque and Angular Momentum

The Physical Setup

Unlocking the Angular Momentum

The Initial State

The Quest for Zero Angular Momentum

The Mathematical Resolution

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r(t)=10αt2i^+5β(t−5)j^​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=m(r×v)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v(t)=dtdr​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L(0)=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L(t)=L(0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r×v=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Cross Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=10 s