# Is there a specific case for the kinetic energy of a particle to be conserved while angular momentum is not conserved?

This is from a question from one of my country's undergraduate entrance exams, so the usual considerations of not taking extraordinary cases apply. The question is, in quotes

"A particle moves with finite velocity and acceleration. The velocity of a particle is perpendicular to its acceleration at any instant of time. Which of the following statements about the particle is always true?

A) Kinetic energy is conserved

B) Acceleration is zero

C) Linear momentum is conserved

D) Angular momentum is conserved."

Only one of the options must be marked correctly. The answer given by their key, says that the kinetic energy must be conserved, but I believe that either both kinetic energy AND angular momentum must be conserved, or neither.

The first task is to note the restraints of the particle.

1. The particle is, in fact, a point particle, and hence cannot rotate about itself, but only about a point other than itself.

2. Velocity $$\vec v$$ and acceleration $$\vec a$$ are finite and perpendicular to each other at all times.

The easiest case that fits the restraints is that of a uniform circular motion of a particle about a center other than itself. In this case $$\vec F$$ is perpendicular to the $$\vec v$$ and so does no work. If $$W = 0$$, there is no change in kinetic energy, and hence it is conserved.

However, $$\vec r$$ and $$\vec F$$ are antiparallel to each other, where $$\vec r$$ describes the position of the particle with respect to the centre. This implies $$\vec r$$ $$\times$$ $$\vec F$$ $$=$$ $$0$$ and hence the external torque $$\vec \Gamma$$ $$=0$$, and the rate of change of angular momentum, $$\frac {d\vec L}{dt}$$ $$=0$$ as well and hence $$\vec L$$ is constant.

To look more generally, for the kinetic energy to be conserved, $$\vec F_{ext}$$ $$=0$$ which either means no forces exist, or all forces are balanced. For angular momentum to NOT be conserved, $$\vec \Gamma_{ext}$$ $$\neq$$ $$0$$. This does not seem possible for a point particle.

For all cases I can think of, angular momentum is more "likely" to be conserved rather than kinetic energy, and when kinetic energy is conserved, the restraints make it so that angular momentum is conserved as well.

In a more obvious sense, since option C) is given incorrect, A) must be incorrect by extension.

I'd like to attach some rigor to my argument though, so if you can explain why exactly I would be wrong, or why they would be wrong, and direct me to some references I can use as proof, that'd be great.

The statement “angular momentum is conserved” implicitly assumes a point about which the angular momentum is to be calculated. For angular momentum to be conserved, any force on the particle must be directed along the line connecting the particle to this central point. It is certainly possible apply a force perpendicular to the particle velocity which is not also along this line, so angular momentum is not in general conserved.

The fact that the force is perpendicular to the velocity does mean that the magnitude of the velocity is constant—only its direction changes. Thus kinetic energy is conserved.

• To add a formula: $\frac{\text{d}}{\text{d}t}\left(\frac{1}{2} m \vec{v}^2\right) = m \vec{a} \cdot \vec{v} = 0$ in the given situation. Commented Jul 19, 2022 at 6:12
• Just for the sake of visual understanding: if I'm orbiting the Earth (perfectly circular orbit), in a way that the Earth is "below me" (at my feet) and I'm facing in the direction of travel; a force acting "downwards" (relative to me) would affect my angular momentum, but a force moving "leftwards" (relative to me) would not affect my angular momentum and therefore be an example of what OP is looking for? Commented Jul 20, 2022 at 9:16

This is same answer as Ben51 but shown in more detail.

The angular momentum of a point particle about a point $$O$$ is $${\bf L} = {\bf r} \times {\bf p}$$ where $${\bf r}$$ is the vector from the $$O$$ to the particle, and $$\bf p$$ is the particle's momentum. Hence $$\frac{d {\bf L}}{dt} = {\bf v} \times {\bf p} + {\bf r} \times \frac{d{\bf p}}{dt} = m {\bf r} \times {\bf a}$$ where we assumed the mass is constant. Hence if $$\bf a$$ is not parallel to $$\bf r$$ then $$\bf L$$ will change. In the problem as stated, $$\bf a$$ is perpendicular to $$\bf v$$ and it is not stated whether or not it is also parallel to $$\bf r$$. In the general case it will not be. Therefore the condition that $$\bf a$$ is perpendicular to $$\bf v$$ is NOT generally sufficient to guarantee that the angular momentum stays constant (but it does guarantee that the kinetic energy will stay constant).

Example For an example, suppose a spaceship is moving along at constant velocity to begin with, and then it fires a steering rocket to the side, giving an acceleration perpendicular to its velocity. Its motion will change direction, its speed will stay constant. If the spaceship was heading directly towards the moon to begin with then its angular momentum about the moon is zero at first, and then non-zero afterwards.

The magnitude of the velocity of the object remains constant, since the acceleration is always perpendicular to the velocity vector - the object changes velocity, but has constant speed. The kinetic energy is therefore constant.

The formula for angular momentum is $$L=mvr$$. We know that the mass and speed are constant, but we cannot guarantee that $$r$$ is constant. It could be, for example in the case of uniform circular motion. But it's not guaranteed to be, it's possible that the object spirals in or out, changing its distance from the center point. Angular momentum may or may not be conserved.

• What I don't understand is: When there is a constant finite acceleration (independent of the direction), how can the velocity be constant? To me the scenario sounds like an "object in orbit", but then the acceleration would be zero. Commented Jul 19, 2022 at 10:02
• @U.Windl The velocity is not constant, only the magnitude of the velocity (speed) is. An object in a circular orbit requires just the right amount of centripetal acceleration, which always points perpendicular to the velocity. An object in orbit does not have zero acceleration, it is accelerating toward the center at all times. It's common to think about acceleration as a change in speed, but that is incorrect - acceleration may affect the speed or the direction of motion. Commented Jul 19, 2022 at 13:39
• But wouldn't that mean that (according to F=m*a) that some energy is constantly added to an object in orbit? Or is it just a continuous lossless conversion between the different kinds of energy (potential vs. kinetic)? So on the long run the sum of velocity and acceleration would be zero? Commented Jul 19, 2022 at 14:01
• @U.Windl A force does no work when applied perpendicular to the direction of motion. W=Fd, but this is the dot product of F and d, which is zero when the two vectors are at right angles. An object in a circular orbit has a constant speed and mass and therefore constant KE, and since it has constant orbital radius, it has constant potential energy as well. An object in a stable orbit will stay there forever, it doesn't require any work to maintain an orbit. Since the centripetal force is always perpendicular to the direction of motion, it does no work. Commented Jul 19, 2022 at 14:51
• OK, I see; I was just confused about "right amount of centripetal acceleration", thinking that the centrifugal acceleration is the same as centripetal acceleration, but with a different sign, so that the effective acceleration would be zero (viewed radius-wise). However the acceleration vectors of centrifugal and centripetal acceleration change all the time, just as the velocity vector does. But still after a full revolution the sum of each would be zero. Somehow for me that would not match the question's "A particle moves with finite velocity and acceleration". Commented Jul 20, 2022 at 9:33

The easiest case that fits the restraints is that of a uniform circular motion of a particle about a center other than itself. In this case $$\vec F$$ is perpendicular to the $$\vec v$$ and so does no work. If $$W = 0$$, there is no change in kinetic energy, and hence it is conserved.

Your analysis is correct except that this is true not just for uniform circular motion but for any type of motion possible with the given conditions. So, (A) is correct.

As for angular momentum, you can exploit the fact that the question says "finite" and not "constant" acceleration. A simple example would be to increase the centripetal acceleration during circular motion which would keep the velocity same but decrease the radius and thus, the angular momentum ($$L=mvr$$). In all of this we didn't violate the condition that acceleration is always perpendicular to velocity. I think this is the case the examiner wants you to think of because the use of the word "finite" over "constant" for acceleration is quite rare in pre-undergrad exams and it seems like a hint.

Besides this, there are several other cases where angular momentum is not conserved( however, I think the above case is the one in focus given my reasoning). Since you consider uniform circular motion, let us continue with that. If you conserve angular momentum about any point(even on the axis) other than the centre of the circle, $$\vec \tau \neq 0$$(for at least some time) since $$\vec \tau= \vec r\times \vec F$$, and $$\vec r$$ is not parallel/antiparallel to $$\vec F$$(for at least some time). Thus, angular momentum is not conserved.

• This is the only answer that gives an explicit and clear example of what the OP asks. Nice one. Commented Jul 20, 2022 at 10:20

My answer is less mathy than those previous, but perhaps more intuitive. When you drop an object such as a coin or playing dice you may observe that when it hits the ground the linear momentum (observed by speed at which it hits the ground) is transferred to angular momentum (observed as it begins spinning quickly if it hits on an edge). The only thing conserved in this case is Kinetic Energy.

• When dropping a physical object, you may find that the kinetic energy is or is not conserved depending on the elasticity of the collision, and you may find that the angular momentum is or is not conserved depending on the angle of collision. I think you need some additional justification to link this scenario to the constraint of "acceleration perpendicular to velocity" to find that KE being conserved is the only thing that's always true. Commented Jul 19, 2022 at 19:36

with Newton equation

$$\frac{d}{dt}\underbrace{(m\,\vec v)}_{\vec p}=\vec F\tag 1$$

thus the linear momentum $$~\vec p~$$ is not conserved

multiply Eq. (1) with $$~\vec v^T~$$

$$\vec v\cdot \frac{d}{dt}(m\,\vec v)=\vec v\cdot\vec F\overset{!}{=}0\\ m\,\int \vec v\cdot\,d\vec v=\int \vec v\cdot F\,dt= \int \frac{\vec {dr}}{dt} \cdot F\,dt=\int\vec F\cdot \vec {dr}=0$$ thus the kinetic energy $$~T=\frac m2\,\vec v\cdot \vec v~$$ is conserved

assume the force $$~\vec F~$$ is constant , thus $$\int\vec F\cdot \vec {dr}=\vec F\cdot\vec r=0~\Rightarrow\vec F\perp\vec r$$

and Euler equation

$$\frac{d}{dt}\underbrace{(I\,\vec \omega)}_{\vec L}=\vec r\times \vec F=\vec\tau\ne \vec 0$$

thus the angular momentum $$~\vec L~$$ is not conserved