Question

A projectile of mass \(m\) is launched from the origin at speed \(v_{0}\) at 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

Answer: The torque acting on the projectile during its flight is \(\vec{\tau} = m(v_0^2 sin(\theta_0)cos(\theta_0) - gt v_0 cos(\theta_0))\hat{k}\).

Step-by-step solution

Setting Up The Motion Equations

The projectile motion can be divided into horizontal and vertical components. We can use the given values and write the equations for horizontal and vertical motions: Horizontal motion: $$ x(t) = v_{0x}t \\ v_{0x} = v_0 cos(\theta_0) $$ Vertical motion: $$ y(t) = v_{0y}t - \frac{1}{2}gt^2 \\ v_{0y} = v_0 sin(\theta_0) $$ Where \(t\) is the time, \(g\) is the acceleration due to gravity, \(v_{0x}\) and \(v_{0y}\) are the horizontal and vertical components of the initial velocity, respectively.

Calculate The Position Vector And Velocity Vector

The position vector \(r\) and velocity vector \(v\) can be written as: $$ \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} \\ \vec{r}(t) = (v_0 cos(\theta_0) t) \hat{i} + (v_0 sin(\theta_0) t - \frac{1}{2}gt^2)\hat{j} $$ $$ \vec{v}(t) = \frac{d\vec{r}(t)}{dt} = v_{0x}\hat{i} + (v_{0y} - gt)\hat{j} \\ \vec{v}(t) = v_0 cos(\theta_0) \hat{i} + (v_0 sin(\theta_0) - gt)\hat{j} $$

Calculate Angular Momentum

Angular momentum \(L\) with respect to the origin is given by: $$ \vec{L} = \vec{r} \times m\vec{v} $$ Using the cross product, we can find the angular momentum: $$ \vec{L} = (v_0 cos(\theta_0) t) \hat{i} \times m(v_0 cos(\theta_0) \hat{i} + (v_0 sin(\theta_0) - gt)\hat{j}) + (v_0 sin(\theta_0) t - \frac{1}{2}gt^2)\hat{j} \times m(v_0 cos(\theta_0) \hat{i} + (v_0 sin(\theta_0) - gt)\hat{j}) $$ The \(\hat{k}\) component of the angular momentum is: $$ L_z = m(v_0 cos(\theta_0) t)(v_0 sin(\theta_0) - gt) - m(v_0 sin(\theta_0) t - \frac{1}{2}gt^2)(v_0 cos(\theta_0)) $$ Simplifying this expression, we get: $$ L_z = m(v_0^2t sin(\theta_0) cos(\theta_0) - \frac{1}{2}gt^2 v_0 cos(\theta_0)) $$ The angular momentum of the projectile about the origin is \(L_z\).

Calculate The Rate Of Change Of Angular Momentum

To find the rate of change of angular momentum, we need to find the derivative of \(L_z\) with respect to time: $$ \frac{dL_z}{dt} = m(v_0^2 sin(\theta_0)cos(\theta_0) - gt v_0 cos(\theta_0)) $$

Calculate The Torque Acting On The Projectile

As there's no external torque acting on the projectile during its motion (negligible air resistance), according to Newton's laws, the rate of change of angular momentum should be equal to the torque \(\vec{\tau}\) acting on the projectile: $$ \vec{\tau} = \frac{dL_z}{dt} $$ So, the torque acting on the projectile during its flight is \(\vec{\tau} = m(v_0^2 sin(\theta_0)cos(\theta_0) - gt v_0 cos(\theta_0))\hat{k}\). In this exercise, we have calculated the angular momentum of a projectile, its rate of change, and the torque acting on it during its flight, using kinematic equations, the definition of angular momentum, and Newton's laws.

Key concepts

Angular Momentum

Angular momentum is a concept in physics that helps describe the motion of spinning or rotating objects. Think of it like the moment when a skater pulls in their arms to spin faster. That's angular momentum at work! When an object moves in a circle or spins around, its angular momentum tells us about how "robust" that spinning motion is.

In the context of projectile motion, even something moving along a path that isn't circular still has angular momentum when measured around a point (like the origin). The angular momentum \(L\) of our projectile is calculated through the cross product of its position vector \(\vec{r}\) and the product of its mass \(m\) and velocity vector \(\vec{v}\). Mathematically, this is given by:

\[ \vec{L} = \vec{r} \times m\vec{v} \]

For the projectile, the important aspect is the \(\hat{k}\) component of \(\vec{L}\), which describes motion out of the plane of motion. Hence, we primarily focus on finding \(L_z\), the component along \(\hat{k}\), providing insight into the twisting motion with respect to the origin.

Understanding angular momentum helps us predict how forces or motions at play might change something spinning. In situations where there's no external torque (force that causes rotation), angular momentum will remain constant. This property is vital in physics, offering insights into systems as simple as projectiles or as complex as entire galaxies.

Torque

Torque is often thought of as "twist" or the rotational equivalent of force. Just like force causes an object to accelerate, torque causes an object to rotate.

In our projectile scenario, we need to find the torque acting on the projectile about the origin. Torque \(\tau\) is linked to the angular momentum by Newton's second law for rotation, which tells us that the rate of change of angular momentum is equal to the torque applied:

\[ \vec{\tau} = \frac{d\vec{L}}{dt} \]

In this exercise, since there is no external force acting in the angular direction (air resistance is negligible), the torque essentially comes from the internal motion components of the projectile. Calculating the torque involves differentiating the angular momentum with respect to time, and hence, we found:

\[ \vec{\tau} = m(v_0^2 \sin(\theta_0)\cos(\theta_0) - gt v_0 \cos(\theta_0))\hat{k} \]

Understanding torque is crucial in many applications, from simple lever systems to complex machinery, ensuring safe and effective operation. For projectiles, knowing the torque helps in understanding how things might spin or rotate around certain points during flight.

Kinematic Equations

Kinematic equations are the bridge between mathematical theory and the physical motion of objects. They help us predict future movement based on current velocity, acceleration, and time.

For any object in motion, these equations break down movements into simpler components to understand complex trajectories better. Projectiles are no different, with both horizontal and vertical motions to consider.

Let's look at our projectile's kinematic equations:

• Horizontal motion: \(x(t) = v_{0x}t\) where \(v_{0x} = v_0 \cos(\theta_0)\)
• Vertical motion: \(y(t) = v_{0y}t - \frac{1}{2}gt^2\) where \(v_{0y} = v_0 \sin(\theta_0)\)

These equations stem from the idea that unless a force acts upon it, an object in motion stays in that motion (Newton's first law).

Kinematics is fundamental in physics and engineering fields allowing for accurate modeling and prediction of motion paths in just about any context you can imagine, from car design to spacecraft navigation.