Question

A projectile of mass \(m\) is launched from the origin at speed \(v_{0}\) and an angle \(\theta_{0}\) above the horizontal. Air resistance is negligible. a) Calculate the angular momentum of the projectile about the origin. b) Calculate the rate of change of this angular momentum. c) Calculate the torque acting on the projectile, about the origin, during its flight.

Short answer

Short Answer Question: Find the angular momentum of a projectile with mass \(m\) launched at an angle \(\theta_0\) with an initial speed of \(v_0\) about the origin, neglecting air resistance. Also, calculate the rate of change of angular momentum and the torque acting on the projectile during its flight. Short Answer: Angular Momentum: \(L(t) = -mv_{0x}(v_{0y}-gt)\hat{k}\) Rate of change of angular momentum: \(\frac{dL}{dt} = m v_{0x}g\hat{k}\) Torque acting on the projectile: \(\vec{\tau} = m v_{0x}g\hat{k}\)

Step-by-step solution

a) Calculate the angular momentum of the projectile about the origin.

First, let's find the components of the initial velocity. The horizontal component is \(v_{0x} = v_{0}\cos\theta_{0}\) and the vertical is \(v_{0y} = v_{0}\sin\theta_{0}\). Consider a time \(t\). At this time, the projectile's position coordinates will be \((x, y) = (v_{0x}t, v_{0y}t - \frac{1}{2}gt^2)\). Here, \(g\) is the acceleration due to gravity and acting vertically downward. The linear momentum of the projectile at time \(t\) is given by \(\vec{p}(t) = m\vec{v}(t) = m(v_{0x}\hat{i} + (v_{0y} - gt)\hat{j})\). The position vector of the projectile with respect to the origin is \(\vec{r}(t) = v_{0x}t\hat{i} + (v_{0y}t - \frac{1}{2}gt^2)\hat{j}\). Now, we can find the angular momentum of the projectile about the origin using the definition of angular momentum, \(L(t) = \vec{r}(t) \times \vec{p}(t)\). Calculating the cross product, we get: $$ L(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_{0x}t & v_{0y}t - \frac{1}{2}gt^2 & 0 \\ mv_{0x} & m(v_{0y} - gt) & 0 \end{vmatrix} $$ Find the determinant and simplifying, we get, $$ L(t) = -m(v_{0x}(v_{0y}-gt))\hat{k} $$ So, the angular momentum of the projectile about the origin is, $$ L(t) = -mv_{0x}(v_{0y}-gt)\hat{k} $$

b) Calculate the rate of change of this angular momentum.

Now we need to find the rate of change of the angular momentum with respect to time. To do this, we differentiate \(L(t)\) with respect to \(t\). $$ \frac{dL}{dt} = -mv_{0x}\left(\frac{d}{dt}(v_{0y}-gt)\right)\hat{k} $$ Since \(v_{0x}\) and \(v_{0y}\) are constant, their derivatives are zero. Differentiating the remaining term with respect to \(t\) we get, $$ \frac{dL}{dt} = -mv_{0x}(-g)\hat{k} $$ So the rate of change of angular momentum is, $$ \frac{dL}{dt} = m v_{0x}g\hat{k}. $$

c) Calculate the torque acting on the projectile, about the origin, during its flight.

From physics, torque \(\vec{\tau}\) is the rate of change of angular momentum, i.e., \(\vec{\tau} = \frac{d\vec{L}}{dt}\). Since the only force acting on the projectile (as air resistance is negligible) is the gravitational force, the torque is caused by this force. From part (b), we found the rate of change of angular momentum: $$ \vec{\tau} = m v_{0x}g\hat{k}. $$ Hence, the torque acting on the projectile about the origin during its flight is \(m v_{0x}g\hat{k}\).

Key concepts

Torque and Angular Momentum

Torque and angular momentum are both fundamental concepts in the realm of rotational dynamics, intimately connected with the motion of bodies in a rotational context. Angular momentum can be thought of as the rotational equivalent of linear momentum, representing the quantity of rotation a body possesses. It is a vector quantity, given by the cross product of the position vector and linear momentum: \[ L = \text{r} \times \text{p} \].

Similarly, torque can be described as the rotational equivalent of force. Torque causes changes in the angular momentum of a body, just as force causes changes in the linear momentum. The mathematical expression for torque is the cross product of the position vector and force: \[ \tau = \text{r} \times \text{F} \]. It is this relationship between torque and angular momentum that governs the rotational dynamics of systems, such as a projectile in motion. In the provided exercise, the need to grasp this relationship becomes apparent as one calculates the torque acting on the projectile through its change in angular momentum.

Projectile Motion Physics

Projectile motion physics provides an analytical approach to predicting the behavior of objects (projectiles) that are launched into the air and subject only to the force of gravity (assuming air resistance is negligible). This sort of motion is characterized by a parabolic trajectory, and it consists of two components: horizontal motion with constant velocity and vertical motion with constant acceleration (due to gravity).

Using the principles of projectile motion, one can deduce the position, velocity, and acceleration of the projectile at any given moment. Important to remember is that horizontal and vertical motions are independent of each other, except for the time aspect. This exercise highlights how projectile motion can be applied to determine the projectile's position and velocity vectors at any time, which are crucial for calculating angular momentum.

Cross Product in Physics

The cross product, also known as the vector product, is a mathematical operation essential in physics, particularly when dealing with vectors in 3-dimensional space. It is primarily used to find a vector that is perpendicular to the plane formed by two other vectors. For vectors \text{A} and \text{B} in three-dimensional space, the cross product is defined as \( \text{A} \times \text{B} = \text{C} \), where \text{C} is a vector perpendicular to both \text{A} and \text{B}.

The direction of vector \text{C} is given by the right-hand rule, and its magnitude is equal to the area of the parallelogram formed by \text{A} and \text{B}. The cross product is pivotal in calculating torque and angular momentum in physics, both of which involve a vector perpendicular to the plane of motion, following the principles applied in the solutions to determine angular momentum of the projectile.

Conservation of Angular Momentum

The conservation of angular momentum is a principle stating that if no external torque acts on a system, the total angular momentum of that system remains constant. In other words, in the absence of external forces or torques, the angular momentum before an event will be equal to the angular momentum after the event. This principle is akin to the conservation of linear momentum but applied to rotating systems.

In an isolated system, or under conditions where external torques can be neglected (such as in our projectile motion scenario where we ignore air resistance), angular momentum is conserved. This concept is typically illustrated in systems such as ice skaters spinning, planets orbiting, and yes, projectiles in flight. The understanding of conservation principles is critical for solving many physics problems and is deeply related to symmetries and fundamental laws of nature.