# Two Dimensional Physics Problems

Two Dimensional Physics Problems

## Inclined Planes

The force in the sliding direction = mg sin θ

Mnemonic: Fsliding direction = mg sLin θ.

The normal force is the other one (mg cos θ).

## Uniform Circular Motion

For any object with an acceleration vector perpendicular to the velocity vector, it will follow a curved path. We can say that the centripetal acceleration of an object in a circular path is: a = v2/r , which means that Fc = mv2/r for all cases of circular motion. |

If the centripetal force is not great enough to fully change the velocity vector to fit back on the path described by a single circle, the object will spiral outwards. If the acceleration is too great for the object’s velocity, it will spiral inwards and crash. The “escape velocity”, then, is exactly the speed that an object needs to stay in the orbit if the only force acting on it is oriented perpendicular to the velocity vector. (Escape velocity is a scalar, so the “velocity” part is a misnomer.) |

This, of course, assumes that the object is unconstrained, and can leave the orbit. Some examples of circular motion involve objects that are constrained to the radius of the path (such as a roller coaster in a loop-de-loop). In these cases, as long as the speed is greater than the escape velocity, it will stay in the circular path (it can’t escape). |

Above: Free-body diagram for a tethered object spinning in a vertical plane. Above: Free body diagram for nonuniform circular motion. The net acceleration vector, a, can be broken into tangential and radial components, aθ and aR = acentripetal, respectively. | Going deeper with the roller coaster example, we can see that it is in fact an example of nonuniform circular motion (where the speed changes). Just based on energy considerations, at the height of the loop, the car has a potential energy of mgh2 compared to the base of the loop. That energy had to come from somewhere, and it came from the kinetic energy the car had when it entered the loop, and since KE = ½ mv2, the velocity of the car is less when it is at the peak of the loop. By looking at the forces involved via the white free-body diagram at left (replace the tension T with Normal Force FN), we can see that the effect of gravity on the “sides” of the loop is to decelerate or accelerate the car tangentially. By solving for accelerations, we see that as the car goes up and down the loop. there is a tangential component to the overall acceleration vector, which is not strictly centripetal anymore. The radial component is still centripetal, and still equals ac=mv2/r. Solving for the tangential component requires calculus, but per our discussion, the tangential component varies with the velocity. So for all cases of circular motion, both uniform and nonuniform, we can say the formula for centripetal acceleration is always the same! |

## Projectile Motion

This graph shows how the angle of launch is related to the distance. The maximum range is when the angle θ = 45o. Notice that for two trajectories, if the offset from that maximum angle, | θ -45 |, is equal, the range is the same.

Also, for all projectile motion problems, just remember these facts:

- The movement in the y direction tells you about the duration of the fall. In other words, the y-direction gives you time.
- If you are ignoring air resistance, there are no forces in the x direction, so vo-x (the initial velocity in the x direction) will determine the range. ∆ x = vo-x t
- Solving difficult problems can be made easier by breaking down the projectile’s arc into two phases of motion:

- Launch until peak
- Peak until touchdown