Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force F = (alpha t i + beta j)

Q: Why is the constant of integration zero when finding velocity and position?

Because the problem states the body is "at rest at the origin at time $t=0$". This means $\mathbf{v}(0) = 0$ and $\mathbf{r}(0) = 0$.

Q: How do I find the magnitude of a vector like displacement?

Use the Pythagorean theorem in 2D or 3D. For a vector $\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}$, the magnitude is $|\mathbf{r}| = \sqrt{x^2 + y^2}$.

Q: Why is the direction of torque important?

Torque is a vector quantity. Its direction tells us the axis of rotation and the sense of rotation (clockwise or counterclockwise) according to the right-hand rule.

Q: What is the right-hand rule for cross products?

If you point your index finger in the direction of the first vector ($\mathbf{r}$) and your middle finger in the direction of the second vector ($\mathbf{F}$), your thumb points in the direction of the cross product ($\mathbf{\tau}$).

Q: Can I just multiply the magnitudes of $\mathbf{r}$ and $\mathbf{F}$ to get torque?

No! Torque is the cross product, so $\mathbf{\tau} = |\mathbf{r}| |\mathbf{F}| \sin(\theta) \hat{\mathbf{n}}$, where $\theta$ is the angle between them. Using the determinant method or component-wise cross product is much safer here.

Question

Consider a body of mass $1.0	ext{ kg}$ at rest at the origin at time $t = 0$. A force $\mathbf{F} = (\alpha t \hat{\mathbf{i}} + \beta \hat{\mathbf{j}})$ is applied on the body, where $\alpha = 1.0	ext{ Ns}^{-1}$ and $\beta = 1.0	ext{ N}$. The torque acting on the body about the origin at time $t = 1.0	ext{ s}$ is $\mathbf{	au}$. Which of the following statements is (are) true ?

Options:

(A) $|\mathbf{	au}| = \frac{1}{3} 	ext{ N-m}$

(B) The torque $\mathbf{	au}$ is in the direction of the unit vector $+\hat{\mathbf{k}}$

(C) The velocity of the body at $t = 1	ext{ s}$ is $\mathbf{v} = \frac{1}{2} (\hat{\mathbf{i}} + 2\hat{\mathbf{j}}) 	ext{ ms}^{-1}$

(D) The magnitude of displacement of the body at $t = 1	ext{ s}$ is $\frac{1}{6} 	ext{ m}$

Sigma Percentile · Accepted Answer

### The Beauty of Kinematics Meets Rotational Dynamics

Welcome to a truly fantastic problem! This question from JEE Advanced 2008 is a beautiful symphony of kinematics and rotational dynamics.

It takes a simple time-varying force and asks us to trace the entire journey of a particle—from its acceleration all the way to the torque it experiences.

I know that seeing time-varying vectors can sometimes feel intimidating, but let's take a breath. We are going to break this down step-by-step, and by the end, you will see just how elegant the physics really is!

### Analyzing the Setup

Let's start by looking at the force acting on our $1.0	ext{ kg}$ body.

We are given the force vector as $\mathbf{F} = (\alpha t \hat{\mathbf{i}} + \beta \hat{\mathbf{j}})$.

Substituting the given values of $\alpha = 1.0$ and $\beta = 1.0$, the force simplifies beautifully to:

$\mathbf{F} = t \hat{\mathbf{i}} + \hat{\mathbf{j}}$

Since the mass is exactly $1.0	ext{ kg}$, Newton's Second Law ($\mathbf{F} = m\mathbf{a}$) tells us that the acceleration vector is numerically identical to the force vector.

So, the acceleration is also:

$\mathbf{a} = t \hat{\mathbf{i}} + \hat{\mathbf{j}}$

### The Master Equation: Finding Velocity

Now, how do we find the velocity? Well, acceleration is simply the rate of change of velocity.

To go from acceleration to velocity, we need to integrate our expression with respect to time.

Integrating $t$ gives us $\frac{t^2}{2}$, and integrating $1$ gives us $t$.

We must also add a constant of integration, $\mathbf{C}$.

$\mathbf{v}(t) = \int (t \hat{\mathbf{i}} + \hat{\mathbf{j}}) dt = \frac{t^2}{2} \hat{\mathbf{i}} + t \hat{\mathbf{j}} + \mathbf{C}$

But wait, the problem states the body starts from rest at time $t = 0$.

This means our constant $\mathbf{C}$ is simply zero.

So, our velocity vector as a function of time is:

$\mathbf{v}(t) = \frac{t^2}{2} \hat{\mathbf{i}} + t \hat{\mathbf{j}}$

Let's check option (c), which asks for the velocity at exactly $t = 1	ext{ s}$.

We just plug $t = 1$ into our velocity equation.

This gives us $\frac{1}{2} \hat{\mathbf{i}} + 1 \hat{\mathbf{j}}$.

If we factor out the $\frac{1}{2}$, we get:

$\mathbf{v}(1) = \frac{1}{2} (\hat{\mathbf{i}} + 2\hat{\mathbf{j}}) 	ext{ ms}^{-1}$

Looking at option (c), we can confidently say it is absolutely correct! One down, three to go.

### Tracing the Path: Finding Position

Moving on, we need the position vector to find the displacement and the torque.

Just like before, velocity is the derivative of position. So, we integrate the velocity vector.

Integrating $\frac{t^2}{2}$ gives $\frac{t^3}{6}$, and integrating $t$ gives $\frac{t^2}{2}$.

Again, the body starts at the origin, so our integration constant is zero.

This leaves us with the position vector $\mathbf{r}$ as a function of time:

$\mathbf{r}(t) = \frac{t^3}{6} \hat{\mathbf{i}} + \frac{t^2}{2} \hat{\mathbf{j}}$

Option (d) talks about the magnitude of displacement at $t = 1	ext{ s}$.

Let's substitute $t = 1$ into our position vector. We get:

$\mathbf{r}(1) = \frac{1}{6} \hat{\mathbf{i}} + \frac{1}{2} \hat{\mathbf{j}}$

To find the magnitude, we take the square root of the sum of the squares of the components.

$|\mathbf{r}(1)| = \sqrt{\left(\frac{1}{6}ight)^2 + \left(\frac{1}{2}ight)^2} = \sqrt{\frac{1}{36} + \frac{1}{4}}$

This simplifies to $\sqrt{\frac{10}{36}}$, which is $\frac{\sqrt{10}}{6}	ext{ m}$.

Option (d) claims it is $\frac{1}{6}	ext{ m}$, so option (d) is incorrect.

**Don't fall for the trap of just looking at the x-component!**

### The Grand Finale: Calculating Torque

Now for the grand finale: the torque!

The torque about the origin is defined as the cross product of the position vector $\mathbf{r}$ and the force vector $\mathbf{F}$.

$\mathbf{	au} = \mathbf{r} 	imes \mathbf{F}$

We need this at $t = 1	ext{ s}$. We already found $\mathbf{r}$ at $t = 1$, and the force $\mathbf{F}$ at $t = 1$ is simply $\hat{\mathbf{i}} + \hat{\mathbf{j}}$.

Let's set up the cross product of these two vectors:

$\mathbf{	au} = \left( \frac{1}{6} \hat{\mathbf{i}} + \frac{1}{2} \hat{\mathbf{j}} ight) 	imes (\hat{\mathbf{i}} + \hat{\mathbf{j}})$

Let's carefully expand this cross product. Remember your cross product rules!

$\hat{\mathbf{i}} 	imes \hat{\mathbf{i}}$ and $\hat{\mathbf{j}} 	imes \hat{\mathbf{j}}$ are zero.

$\hat{\mathbf{i}} 	imes \hat{\mathbf{j}}$ is $\hat{\mathbf{k}}$, and $\hat{\mathbf{j}} 	imes \hat{\mathbf{i}}$ is $-\hat{\mathbf{k}}$.

So, we get:

$\mathbf{	au} = \frac{1}{6} \hat{\mathbf{k}} - \frac{1}{2} \hat{\mathbf{k}}$

Finding a common denominator, this becomes $\frac{1 - 3}{6} \hat{\mathbf{k}}$, which simplifies beautifully to:

$\mathbf{	au} = -\frac{1}{3} \hat{\mathbf{k}} 	ext{ N-m}$

**Notice the minus sign!** It means the torque is directed into the page, along the negative z-axis.

### Final Conclusion

Finally, let's check options (a) and (b).

The magnitude of our torque is exactly $\frac{1}{3}	ext{ N-m}$. So, option (a) is perfectly correct!

However, the direction is $-\hat{\mathbf{k}}$, not $+\hat{\mathbf{k}}$. So option (b) is incorrect.

The negative sign is a classic trap! Always pay attention to the order in the cross product.

So, our final correct answers are options (a) and (c). Keep practicing, and you'll master these concepts in no time!

Select Answer:

Visualized Solution

$Force and Acceleration$

$Velocity Vector$

$Checking Option (c)$

$Position Vector$

$Checking Option (d)$

$Calculating Torque$

$Evaluating the Cross Product$

$Checking Options (a) and (b)$

The Sigma Insight: Torque and Angular Momentum

The Beauty of Kinematics Meets Rotational Dynamics

Analyzing the Setup

The Master Equation

Tracing the Path

The Grand Finale

Final Conclusion

Similar Questions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force and Acceleration

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Options (a) and (b)