Circular Motion and Gravitation

Why don't planets fly off into space?
Why don't cars turn without sliding off a curb?
What is the mechanics between a yoyo and how does it relate to circular motion gravitation?
The answer lies in the invisible forces of centripetal acceleration.

In the case of a pendulum, the object is constantly changing direction as it goes around in a circle.
Velocity is a vector quantity, meaning it has both direction and magnitude.
Acceleration is a change in velocity.
Any change in direction over time is acceleration.
Centripetal acceleration: The acceleration of an object moving in a circle.

a_c = \frac{v^2}{r}
- a_c: centripetal acceleration
- v: velocity
- r: radius of the circle

Centripetal force is the net force causing circular motion.
Centripetal force is not a force on its own; it is provided by tension, friction, gravity, or normal force, depending on the situation.
The tension force is pulling inwards towards the center of the circle and mg is pulling down on the yoyo.
F_t is the net force pulling inwards towards the center of the circle creating its centripetal acceleration.
F = ma can be rewritten as Ft = mac where a_c = \frac{v^2}{r}.
F_t = \frac{mv^2}{r}
If the radius decreases over time, and the tension force is equal and constant, the velocity must increase.

A roller coaster traveling along a loop is an example of circular motion.
At every point along the track, there is centripetal acceleration going towards the center of the loop.
If the loop de loop was taller, and the cart started at rest at a certain height, all the energy would be converted back into gravitational potential energy at the top of the loop, leaving no kinetic energy (velocity).
If there is no velocity, then there is no centripetal acceleration (a_c = \frac{v^2}{r} = 0), and the car would fall.

Momentum is always conserved.
Li = Lf
L = I\omega
- L: angular momentum
- I: moment of inertia
- \omega: angular speed (\omega = \frac{v}{r})
I = mr^2 for a point mass
Speed is greater when the radius is smaller, and vice versa, so angular momentum stays unchanged.

Fg = mac
v = \frac{d}{t}
t is the period (T), the amount of time it takes to travel one revolution.
Kepler's Law: T^2 = \frac{4\pi^2 r^3}{GM}
- T: period
- r: radius
- G: universal gravitational constant
- M: mass it's rotating around
T^2 is directly proportional to r^3, linking centripetal acceleration and gravitation.

At the bottom, tension and gravity act toward the center, so tension is greatest.
- T_{bottom} = \frac{mv^2}{r} + mg
At the top, gravity already pulls inward, so tension drops to its minimum.
- T_{top} = \frac{mv^2}{r} - mg