Question

At time \(t=0\), a \(2 \mathrm{~kg}\) particle has position vector \(\vec{r}=(4 i-2 j) m\) relative to the origin. Its velocity is given by \(\overrightarrow{\mathrm{v}}=2 \mathrm{t}^{2} \overline{\mathrm{i}}(\mathrm{m} / \mathrm{s})\) for \(\mathrm{t}\) \(\geq 0\). The torque acting on the particle about the origin at \(t=2 s\), is : (1) \(32 \mathrm{~N}-\mathrm{m}(\hat{\mathrm{K}})\) (2) \(16 \mathrm{~N}-\mathrm{m}(\hat{\mathrm{K}})\) (3) \(16 \mathrm{~N}-\mathrm{m}(\hat{\mathrm{K}})\) (4) \(12 \mathrm{~N}-\mathrm{m}(\hat{\mathrm{K}})\)

Short answer

The torque acting on the particle at \( t = 2 \) s is \( 32 \hat{k} \) N·m.

Step-by-step solution

- Identify Initial Position and Velocity

At time \(t=0\), the particle has a position vector \( \vec{r} = (4 \hat{i} - 2 \hat{j}) \, m \). The velocity of the particle is given by \( \vec{v} = 2 t^2 \hat{i} \) m/s for \(t \geq 0\).

- Determine Position Vector at \( t = 2 \) s

To find the position vector \( \vec{r} \) at \( t = 2 \) s, note that the velocity \( \vec{v} \) contributes only in the \( \hat{i} \) direction. Thus, \(\vec{r}(2) = (4 + \int_0^2 2t^2 dt) \hat{i} - 2 \hat{j} \). Integrating the velocity term, we get: \[\int_0^2 2t^2 dt = [\frac{2t^3}{3}]_0^2 = \frac{2(2^3)}{3} = \frac{16}{3} \text{ m} \]. Therefore, \(\vec{r}(2) = (4 + \frac{16}{3}) \hat{i} - 2 \hat{j} = \frac{28}{3} \hat{i} - 2 \hat{j} \).

- Calculate Linear Momentum at \( t = 2 \) s

The linear momentum \( \vec{p} \) is given by \( \vec{p} = m \vec{v} \). At \( t = 2 \) s, \( \vec{v} = 2 (2^2) \hat{i} = 8 \hat{i} \) m/s, so \(\vec{p} = 2 \times 8 \hat{i} = 16 \hat{i} \) kg m/s.

- Determine Torque Using Cross Product

Torque \( \vec{\tau} \) is given by the cross product \( \vec{\tau} = \vec{r} \times \vec{p} \). Here, we use \( \vec{r} = \frac{28}{3} \hat{i} - 2 \hat{j} \) and \( \vec{p} = 16 \hat{i} \). Thus: \[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \frac{28}{3} & -2 & 0 \ 16 & 0 & 0 \end{vmatrix} = -16(-2) \hat{k} = 32 \hat{k} \, \text{N} \cdot \text{m} \]

Key concepts

torque calculation

Torque is a measure of how much a force acting on an object causes that object to rotate. The point where an object pivots is called the axis. To calculate torque, we use the formula: \(\vec{\tau} = \vec{r} \times \vec{F}\). In this formula, \( \vec{\tau} \) represents the torque vector, \( \vec{r} \) is the position vector, and \( \vec{F} \) is the force vector. The symbol \( \times \) denotes the cross product, which is a way to multiply two vectors that results in a vector quantity. An important aspect of torque is its direction, which is given by the Right-Hand Rule. In problems involving particles, like the one in the exercise, the torque is often calculated using the particle's position and its linear momentum. We further utilize the modified formula: \( \vec{\tau} = \vec{r} \times \vec{p} \), where \( \vec{p} \) is the linear momentum of the particle.

position vector

The position vector gives the location of a particle relative to a reference point, often the origin. In this exercise, the position vector at time \( t = 0 \) is given as \(\vec{r}(0) = 4 \hat{i} - 2 \hat{j} \) meters. Over time, this position changes due to the particle's velocity. To find the new position at a specific time, we integrate the given velocity over that time. For example, for a velocity \( \vec{v} = 2t^2 \hat{i} \), the change in position in the \( \hat{i} \) direction can be determined by integrating the velocity function: \[ \int_0^2 2t^2 dt = \frac{2t^3}{3}\Big|_0^2 \]. Solving the integral gives the change in the \(\hat{i} \) component. Adding this to the initial position, you get the new position vector at \( t = 2 \) seconds: \(\vec{r}(2) = \left(4 + \frac{16}{3}\right) \, \hat{i} - 2 \hat{j} = \frac{28}{3} \hat{i} - 2 \hat{j} \). Understanding how the position vector evolves over time is crucial for calculating subsequent physical quantities like torque.

linear momentum

Linear momentum is an essential concept in physics, representing the product of a particle's mass and its velocity: \( \vec{p} = m \vec{v} \). It is a vector quantity, meaning it has both a magnitude and a direction. For the particle described in the exercise, with mass \(m = 2 \) kg and velocity \(\vec{v} = 2t^2 \hat{i} \) m/s, we can find the linear momentum at a certain time by substituting the velocity at that time into the equation. For example, at \( t = 2 \) seconds, the velocity is \( 8 \hat{i} \) m/s, making the linear momentum \( \vec{p} = 2 \times 8 \hat{i} = 16 \hat{i} \) kg·m/s. Once the linear momentum is known, it can be used in various calculations, such as finding the torque acting on the particle.