Question

The position vector \(\vec{r}\) of a particle of mass \(m\) is given by the following equation $$ \vec{r}(t)=\alpha t^{3} \hat{i}+\beta t^{2} \hat{j}, $$ where \(\alpha=10 / 3 \mathrm{~m} \mathrm{~s}^{-3}, \beta=5 \mathrm{~m} \mathrm{~s}^{-2}\) and \(m=0.1 \mathrm{~kg} .\) At \(t=1 \mathrm{~s}\), which of the following statement(s) is(are) true about the particle? (A) The velocity \(\vec{v}\) is given by \(\vec{v}=(10 \hat{i}+10 \hat{j}) \mathrm{ms}^{-1}\) (B) The angular momentum \(\vec{L}\) with respect to the origin is given by \(\vec{L}=-(5 / 3) \hat{k} \mathrm{~N} \mathrm{~m} \mathrm{~s}\) (C) The force \(\vec{F}\) is given by \(\vec{F}=(\hat{i}+2 \hat{j}) \mathrm{N}\) (D) The torque \(\vec{t}\) with respect to the origin is given by \(\vec{t}=-(20 / 3) \hat{k} \mathrm{Nm}\)

Short answer

At \(t=1\) s, (A) The velocity \(\vec{v}\) is correctly given by \(10 \hat{i} + 10 \hat{j}\) m/s, (B) The angular momentum \(\vec{L}\) is indeed \(-\frac{5}{3} \hat{k}\) Nm\cdot s, (C) The force \(\vec{F}\) is correctly expressed by \(\hat{i} + 2 \hat{j}\) N, and (D) The torque \(\vec{\tau}\) is \(-\frac{20}{3} \hat{k}\) Nm.

Step-by-step solution

Calculate the velocity

To calculate the velocity \(\vec{v}(t)\), take the first derivative of the position vector \(\vec{r}(t)\) with respect to time \(t\). Using the given equation \(\vec{r}(t) = \alpha t^{3} \hat{i} + \beta t^{2} \hat{j}\), where \(\alpha = \frac{10}{3} \mathrm{m/s^{3}}\) and \(\beta = 5 \mathrm{m/s^{2}}\), the derivative is \(\vec{v}(t) = \frac{d}{dt}(\alpha t^{3} \hat{i} + \beta t^{2} \hat{j}) = 3\alpha t^{2} \hat{i} + 2\beta t \hat{j}\). Substituting \(\alpha\) and \(\beta\) values and evaluating at \(t=1\) s, the velocity becomes \(\vec{v}(1) = 3\times\frac{10}{3} \times 1^{2} \hat{i} + 2\times 5 \times 1 \hat{j} = 10 \hat{i} + 10 \hat{j}\) m/s.

Calculate the angular momentum

The angular momentum \(\vec{L}\) of a particle with respect to the origin is given by the cross product of the position vector \(\vec{r}\) and the momentum \(\vec{p} = m\vec{v}\). Using \(\vec{r}(1) = \alpha \times 1^{3}\hat{i} + \beta \times 1^{2}\hat{j}\) and knowing that the momentum \(\vec{p}(1) = m\vec{v}(1)\), we have \(\vec{L}(1) = \vec{r}(1) \times \vec{p}(1) = (\alpha \hat{i} + \beta \hat{j}) \times (m\cdot 10 \hat{i} + m\cdot 10 \hat{j})\). Substituting \(\alpha = \frac{10}{3}\), \(\beta = 5\), and \(m = 0.1 \mathrm{kg}\) into the expression and using the right-hand rule for the cross product gives \(\vec{L}(1) = -\frac{5}{3} \hat{k} \mathrm{Nm\cdot s}\).

Calculate the force

The force \(\vec{F}\) acting on the particle can be found using Newton's second law, \(\vec{F} = m\vec{a}\), where \(\vec{a}\) is the acceleration of the particle. The acceleration is the second derivative of the position vector with respect to time. Starting with \(\vec{v}(t) = 3\alpha t^{2} \hat{i} + 2\beta t \hat{j}\), take the derivative with respect to \(t\) to get \(\vec{a}(t) = \frac{d}{dt}\vec{v}(t) = 6\alpha t \hat{i} + 2\beta \hat{j}\). Evaluate it at \(t = 1\) s: \(\vec{a}(1) = 6\alpha \hat{i} + 2\beta \hat{j}\). Substituting values of \(\alpha, \beta\), and \(m\) gives the force as \(\vec{F}(1) = m(6\alpha \hat{i} + 2\beta \hat{j}) = 0.1(\frac{60}{3}\hat{i} + 10\hat{j}) = (\hat{i} + 2\hat{j}) \mathrm{N}\).

Calculate the torque

Torque \(\vec{\tau}\) with respect to the origin is found by taking the cross product of the position vector \(\vec{r}\) and the force \(\vec{F}\). From the previous steps, using \(\vec{r}(1) = \alpha \hat{i} + \beta \hat{j}\) and \(\vec{F}(1) = \hat{i} + 2\hat{j}\), calculate \(\vec{\tau}(1) = \vec{r}(1) \times \vec{F}(1) = (\alpha \hat{i} + \beta \hat{j}) \times (\hat{i} + 2\hat{j})\). Substitute the values for \(\alpha\) and \(\beta\), and use the right-hand rule to find \(\vec{\tau}(1) = -(\frac{20}{3} \hat{k}) \mathrm{Nm}\).

Key concepts

Angular Momentum

Angular momentum is a fundamental concept in physics, representing the rotational inertia of an object in motion about a point. It can be envisioned as the rotational equivalent of linear momentum and is crucial in the study of systems involving rotational dynamics.

In mathematical terms, the angular momentum \( \vec{L} \) of a particle of mass \(m\) with a position vector \( \vec{r} \) and linear momentum \( \vec{p} \) is the cross product of \( \vec{r} \) and \( \vec{p} \), expressed as \( \vec{L} = \vec{r} \times \vec{p} \). For our exercise involving the particle with a mass of 0.1 kg and the specific position vector given, the angular momentum with respect to the origin is calculated at time \( t = 1 \) second. The careful evaluation of the cross product between the position vector and the particle's momentum at that instance leads to the correct value of \( \vec{L} \).

Given that the calculation of angular momentum involves the cross product, let's delve into the intricacies of what a cross product entails and how it applies to our problem.

Cross Product

The cross product, a key operation in vector algebra, is utilized when calculating quantities like angular momentum and torque. It gathers two vectors and results in a third vector that is perpendicular to the plane formed by the initial vectors, following the right-hand rule for direction.

The cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is denoted as \( \vec{A} \times \vec{B} \). In terms of components, if \( \vec{A} \) has components \( A_x \), \( A_y \) and \( A_z \) and \( \vec{B} \) has components \( B_x \), \( B_y \) and \( B_z \), the cross product's components along the x, y and z axes are determined by a determinant involving these components. In the context of our physics problem with the particle, we applied the cross product to find both the angular momentum and the torque. The operation reveals the direction and magnitude of these rotational vectors based on the particle's positional and force vectors at the given time.

Physics Problem Solving

Physics problem solving is an intellectual exercise that requires a step-by-step approach to untangle complex concepts and apply mathematical principles to physical scenarios. The key steps typically involve identifying the known and unknown variables, understanding the physical principles involved—such as Newton's laws, conservation of energy or momentum—and applying the appropriate mathematical techniques.

For a systematic approach to problem solving, one may start by sketching diagrams to visualize the situation, writing down the relevant equations, and then progressively solving for the quantities of interest. In our example, we systematically derived the velocity, angular momentum, force, and torque for a particle given its position vector and mass. Problem solving in physics is not just about plugging numbers into equations; it also encompasses a deep understanding of the interplay between physical quantities and the use of critical thinking to select the appropriate concept or law to apply in each scenario.

Let's connect this approach with the specific law that was used to calculate the force on the particle—Newton's second law.

Newton's Second Law

Newton's second law of motion is one of the most foundational principles in classical mechanics, serving as the cornerstone for dynamics. It can be succinctly stated: The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The law is encapsulated by the famous equation \( \vec{F} = m\vec{a} \), where \( \vec{F} \) is the net force applied to an object, \(m\) is its mass, and \(\vec{a}\) is its acceleration.

Applying this principle to the given physics problem, we find the force on the particle by differentiating its velocity—a process providing the acceleration—and then multiplying by the particle's mass. It's crucial to grasp that the acceleration is not constant but varies with time, as indicated by the particle's time-dependent position vector. Thus, Newton's second law not only offers a way to determine the force but also ties together the concepts of acceleration and mass, revealing the intricate dance between forces and motion in our universe.