Example
- At every instant, the acceleration deflects the satellite into toward the center, but the velocity moves away. The result is a circular path around the center.
March 11, 2019
Prior Skills:
Know how to decompose position, velocity, and acceleration vectors into component vectors.
Know how to take component vectors of position, velocity, and accelerations to combine to a resultant vector.
Know how acceleration affects both the magnitude and direction of velocity.
Be able to apply these to practical problems.
Learning Goals:
Define and Identify the Centripetal Acceleration and Tangential Acceleration in Circular Motion.
Know how to calculate rotational speed, acceleration, and period in circular motion.
Be able to apply uniform circular motion to practical problems.
When velocity is in the same direction as acceleration an object will increase in speed in that direction.
When velocity is in the opposite direction as acceleration an object will decrease in speed in that direction.
Vectors in the opposite direction are said to be anti-parallel.
If the object accelerates long enough it will instantaneously stop, reverse direction and speed up in the direction of acceleration.
If there is nothing to stop the object in motion it will instantaneously stop then start speeding up in the direction of acceleration, like a ball thrown upward.
Note: Just because a velocity vector is negative, that doesn’t necessarily mean the object is moving backward. Most of the time it’s headed forward in the opposite of the positive direction.
When an acceleration vector is perpendicular to velocity, this results in a change in direction.
This change in direction will follow a curved path: Parabolic, Hyperbolic, Elliptical, or Circular depending on the exact magnitudes and directions of the velocity and acceleration vectors.
The acceleration vector is always directed to the inside of the curved path, e.g., toward the center of a circular trajectory.
We will focus on when velocity and acceleration are exactly \(90^o\), which will result in Uniform Circular Motion (UCM).
https://phet.colorado.edu/sims/my-solar-system/my-solar-system_en.html
Note that for solar system objects the paths are not exactly circles.
As seen, when there is a component of acceleration that is 90\(^o\) to the velocity, that component of acceleration does not speed the object up or slow it down. It changes the object’s direction.
The curved path will follow a circle or circular arc.
If the acceleration is exactly 90\(^o\), only direction will change, speed will stay the same.
This is called \(\textbf{Uniform Circular Motion}\).
Non-uniform Circular motion occurs when the angle between the velocity vector and acceleration vector is not exactly 90\(^o\). This will be covered at a later time.
If the angle is less than 90, the object speeds up while moving in a circle.
If the angle is greater than 90, the object slows down while moving in a circle.
In both Uniform Circular Motion and Projectile Motion, the magnitude of acceleration is constant.
However, In Uniform Circular Motion (UCM), the direction of the acceleration vector changes.
The direction of acceleration in UCM is always directed to the center of the curved path.
In Projectile Motion, the direction of the acceleration does not change; it is directed to the surface of the Earth.
Radial Acceleration: Acceleration component 90\(^o\) from velocity.
Always Directed toward inside of circular path.
This is the Centripetal Acceleration in circular motion.
Tangential Acceleration: Acceleration component (anti-)parallel from velocity.
Period: The duration of time to travel one complete circle
Measured in [time] units: seconds.
Will be constant in UCM.
Rotational Speed: The distance traveled along the circumference of the circular path with respect to time.
Measured in speed units [length]/[time]: m/s
Will be constant in UCM
$$
F_c = m \
v =
$$
Where \(F_c\) is centripetal Force, m is mass, v is rotational speed, R is radius of path and T is period.
Therefore,
$$
a_c = =
$$
Where a\(_r\) is radial acceleration.
Assuming a circular orbit, what is the radial acceleration of the Earth as it orbits the Sun? The Period of Earth’s orbit is 365.25 days and the radius of the orbit is \(1.496*10^{11} m\).
$$
T = 365.25 d * * = 3.155810^7 s; R = 1.49610^{11} m
$$
$$
a_r = =
$$
$$
a_r = = 5.930*10^{-3}
$$
Saturday Morning Breakfast Cereal by Zach Weinersmith.
An RC car is moving in UCM. It travels at a speed of 5.00 m/s and completes a circle every 15.71 s. What is the radius of its path?
$$
v = \
R = \
R =
R = 12.5 m
$$
In many science fiction books, movies, and TV shows, artificial gravity is achieved by using centripetal acceleration. If a space station maintains Earth’s gravitational acceleration on a deck that is a radius of 100 m away from the axis of rotation, then what is the period of rotation? What is the rotational speed?
$$
a_c = 9.8 \
R = 100 m \
a_c = = \
T = = \
\
v= = =
$$
Banking a Curve is when a road or track is titled in the dimension perpendicular to the velocity of the vehicles making the turn.
In an unbanked turn, only friction acts as centripetal force for the turn.
Recall that static friction has an upper limit.
If static friction is exceeded, the car will slide in a straight line, risking collisions or running off the road.
Banking a turn makes the normal force acting on the car provide the centripetal force.
Normal force is also unaffected by rain, ice or bald tires.
Assume the road has a banked curve at angle, \(\beta\) from the vertical, and radius \(R\).
The vertical component of Normal\(F_n,y\) has to equal weight, \(mg\).
$$
F_{n,y} = F_n*cos() \
F_n =
$$
$$
F_{n,x} = F_nsin() = mg = mgtan() \
m = mg*tan() v =
$$
Where v is the speed of the turn that will not need friction.
$$
m = _smg \
v =
$$
Note that tires on dry asphalt have static coefficients of friction that measure between 0.7 and 1. This is similar to banked turns of 35o to 45o.
The danger being \(\mu_s\) depends on conditions of the road and tires.
The NASCAR track at Talladega, AL has a turn with a bank angle of 33o and turn radius of 1000 ft. What is the maximum speed a driver can take the turn and not have to rely on traction to keep them on the track?
$$
v = = \
144 = 98
$$
Do you think that Normal force accounts for all the centripetal force?
The lift force of a plane, which balances out weight during steady flight, provides the centripetal force when an airplane turns. This is why airplanes always tilt when they turn in flight.
What is the bank angle for a plane turning at 200. km/hr with a turn radius of 500. m?
$$
g*tan() = = tan^{-1}() \
200 = 55.6 \
= tan^{-1}() = 32.2^o
$$