I was riding my bicycle downhill. When the street leveled I realized I had forgotten something so I started turning around without pedaling, pretty much using the momentum I had acquired. Soon I was moving in the complete opposite direction, still without having to pedal.

I couldn't help but thinking how that was possible.

I understand the momentum has a direction arrow and somehow I changed that arrow to the opposite direction. Of course I was going much slower by the time I completely turned around, but still moving.

  • Is the friction of the wheel against the floor somehow like tiny collisions that "bounce" that momentum force gradually changing that direction? (To the expense of losing energy quicker, thus, speed)?

  • I used the bike example only because that's when the doubt popped up, but the same I can ask about a ball rolling down and going through a loop or a skier going down a hill and jumping up in a 90 degrees (relative to the downhill part) ramp

What's the physics and maths behind the momentum change to the opposite or perpendicular direction of a moving object when no extra force is applied?


3 Answers 3


The change in the linear momentum of an object requires a net force to act on the object. A change in the angular momentum of an object requires a net torque to act on the object. So the answer to your question "What's the physics and math behind the momentum change to the opposite or perpendicular direction of a moving object when no extra force is applied?" is: linear (angular) momentum cannot change without a net force (torque).

Also, do not confuse change in momentum with work. Momentum is a vector and work is a scalar. For example, if the force is perpendicular to the direction of motion the force does no work, but can change the linear momentum. An example is an object spinning on a string in a circle. The tension in the string provides the centripetal force, but this force is perpendicular to the circular motion. The tension changes the linear momentum by changing the direction of the velocity vector but the speed, hence the kinetic energy, is constant.

For the ball/skier, the force of constraint of the ground changes direction as the loop bends; for the bike, the force of friction changes direction as you turn the wheel, and if you lean over the torque from gravity comes into play.


With no forces, the bicycle will travel in a straight line at constant speed. If a forces is applied forward, it will follow the same direction but speed up. If backward, it will slow down.

A sideways force is neither forward nor backward. It will neither speed the bike up, nor slow it down. It will change the direction.

There are of course forces with components in the direction of motion and perpendicular to it. The effect of such a force is the same as each force applied separately.

You can see this from energy as well.

$$P = \vec{F} \cdot \vec{v}$$

If the angle between $\vec{F}$ and $\vec{v}$ is less than $\pi/2$, then $P > 0$ and the bike gains kinetic energy. If the angle between $\vec{F}$ and $\vec{v}$ is greater than $\pi/2$, then $P < 0$, and the bike loses kinetic energy. If the angle is exactly $\pi/2$, then $\vec{F} \cdot \vec{v} = P = 0$, and the kinetic energy stays constant.

Bicycles are designed so when you lean, the front wheel turns. This steers you in a circle. If you pedal hard enough to overcome friction, the net forward/backward force is $0$, and your speed stays constant.

If you go in a circle at constant speed, there is always a sideways centripetal force. In this case, the source is static friction from the wheel's contact with the street. (You couldn't turn on ice, where there is no friction.)

If you don't pedal as you turn, the sideways force still causes you to turn. But now there is also a net backward force that slows you as you proceed.

  • $\begingroup$ Why was this down-voted? It is correct that a net force is required to change momentum. $\endgroup$
    – John Darby
    Commented Dec 1, 2020 at 17:48
  • $\begingroup$ Perhaps because there was a mistake: "If the angle =π/2, P⋅v=0, and the kinetic energy stays constant." Should be be $F$ not $P$. Also vector symbols would add clarity I think. $\endgroup$
    – user256872
    Commented Dec 1, 2020 at 22:40
  • $\begingroup$ Yes, thanks for the comment. $\endgroup$
    – John Darby
    Commented Dec 1, 2020 at 23:14
  • $\begingroup$ Thank you for pointing that out and fixing it. $\endgroup$
    – mmesser314
    Commented Dec 1, 2020 at 23:58

Your momentum is not conserved because there are external forces acting on the you + bicycle system. Though there are several external forces, the main one is friction.

The friction from the ground will will provide a torque and force on your bicycle, hence your angular momentum will not be conserved.


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