Chapter 11 - Projectiles and Circular Motion

Chapter 11 (Unit 5)

Projectiles and Circular Motion

Projectile Motion A projectile is any object that has been given

an initial thrust and then allowed to soar through the air under the force of gravity.

The range of the projectile is the horizontal distance that it travels.

Projectile Motion What do we already know about projectiles?

Well, let’s use a reference point: something we all know about either because we’ve done it or heard about it.

Projectile Motion and Angry Birds How can you get the bird to travel farthest?

Are there multiple angles that will land you in the same spot?

What happens if you launch the birds at the same angle but with different initial velocities?

Projectile Motion and Angry Birds What happens when you launch the birds at

the same velocity but different angles?

What launch angles have the longest time in flight?

What launch angles have the shortest time in flight?

Projectile Motion What sorts of things might affect the

trajectory of a projectile?

Projectile Motion Fortunately for you, the mathematics needed

to account for these is extremely complex, so we will not be dealing with them.

Even better! You don’t need to learn any new concepts to analyze and predict the motion of projectiles!

All you need are data that will provide you with the velocity of the projectile at the moment it was launched, and the kinematic equations for uniformly accelerated motion.

Projectile Motion Do you think the vertical and horizontal

motions of the projectile are dependant or independent of each other?

Projectile Motion Let’s try it and see!!

Projectile Motion So, projectiles’ horizontal and vertical motions

are independent of each other!

Some important things to remember: Gravity is the only force influencing our “ideal”

projectiles Gravity ONLY affects the vertical motion, so the

equations for uniformly accelerated motion apply No forces affect horizontal motion, so the equations for

uniform motion apply

Projectile Motion So, if we have two independent equations,

how can we connect them together?

Projectile Motion So, what are the equations for uniformly

accelerated motion that we can use? (vertical component)

What are the equations for uniform motion that we can use? (horizontal component)

Model ProblemWhile hiking in the wilderness, you come to a

cliff overlooking a river. A topographical map shows that the cliff is 291m high and the river is 68.5m wide at that point. You throw the rock directly forward from the top of the cliff, giving the rock a horizontal velocity of 12.8m/s

a. Did the rock make it across the river?b. With what velocity did the rock hit the

ground or water?

Model Problem

Model Problem

Projectiles Launched at an Angle Most projectiles do not start their trajectory

horizontally – rather, they start at an upward angle to the horizontal.

Because of this, they have an initial velocity in BOTH the horizontal and vertical directions.

These trajectories are described as being parabolas.

Projectiles Launched at an Angle The only extra step that we need in order to

analyze a parabolic projectile is to determine the magnitude of the horizontal and vertical components of the initial velocity.

So, now we will be sure to include the “vy,i” in our equations for the vertical motion of the projectile.

Model ProblemA golfer hits the golf ball off the tee, giving it an

initial velocity of 32.6m/s at an angle of 65º from the horizontal. The green where the golf ball lands is 6.30m higher than the tee, as shown below. Find:

a. The time interval during which

the golf ball was in the air.

b. The horizontal distance it travelled.

c. The velocity of the ball just before

hit the ground.

Model Problem

Model Problem

Model Problem

Symmetrical Trajectories If a projectile lands at the same level it was

launched at, then the trajectory is a perfect parabola.

If this is the case, then we can simplify many of our equations.

Symmetrical TrajectoriesTime of Flight:

Symmetrical TrajectoriesRange:

Symmetrical TrajectoriesMaximum Height:

Model ProblemA player kicks a football for the opening kickoff.

He gives the ball an initial velocity of 29m/s at an angle of 69º from the horizontal. Neglecting friction, determine the ball’s maximum height, hang time, and range.

Model Problem

Uniform Circular Motion Have you ever been on a ride at an

amusement park that spins really fast and goes upside down?

Why do you feel “stuck” against the wall? What is keeping you there?

What would happen if the ride stopped turning while it was in the air?

Centripetal Acceleration When an object is moving in a circle and its

speed (the magnitude of the velocity) is constant, it is said to be moving with uniform circular motion.

The direction of the object’s velocity is always tangent to the circle:

Centripetal Acceleration Since the direction of motion is always

changing, the object is always accelerating.

So, how can we describe the acceleration for triangle OPQ?

Centripetal Acceleration We need to show that OPQ is similar to the triangle

formed by the velocity vectors:r1 = r2 because they are the radii of the same circle

|v1| = |v2| because the speed is constant

r1 is perpendicular to v1 and r1 is perpendicular to v2

θr = θv

Based on all of these, the two isosceles triangles are similar

Centripetal Acceleration Now, we can use the similar triangles to find

the magnitude of the acceleration.

Centripetal Acceleration

Centripetal Acceleration Trying to describe the acceleration in the x and y

coordinate system would be very difficult because the direction is always changing.

Instead, we talk about the magnitude of the acceleration (which is constant) and just say that the direction is always towards the center of the circle.

So, we call it the centripetal acceleration, ac , so that it only accounts for the magnitude – it is not a vector.

Centripetal Force According to Newton’s laws of motion, what

causes an object to accelerate?

Since an object moving with uniform circular motion is always accelerating, there must always be an exerted force acting on it.

This force is in the same direction as the acceleration.

Centripetal Force Because the force causing centripetal

acceleration is always pointed towards the center of the circular path, it is called the centripetal force.

This force is extremely different than any other forces you’ve talked about!

It is not a force like gravity or friction – it is a force that is required for an object to move circularly.

Centripetal Force This force can be supplied by any type of force

(e.g. gravity, friction, tension, etc).

Two forces can also create a centripetal force, instead of just the one (a combination of forces)

Centripetal Force We can find the magnitude of the centripetal

force required to cause an object to travel in a circular path by applying Newton’s second law to a mass moving with centripetal acceleration.

Centripetal Force

Model ProblemA car with a mass of 2135kg is rounding a curve

on a level road. If the radius of curvature of the road is 52m and the coefficient of friction between the tires and the road is 0.70, what is the maximum speed at which the car can make the curve without skidding off the road?

Model Problem

Model ProblemYou are playing with a yo-yo with a mass of

225g. The full length of the string is 1.2m. You decide to see how slowly you can swing it in a vertical circle while keeping the string fully extended, even when the yo-yo is at the top of its swing.(a) Calculate the minimum speed at which you can swing the yo-yo while keeping it on a circular path.

(b) At the speed that you determine in part a, find the tension in the string when the yo-yo is at the side and at the bottom of its swing.

Model Problem

Model Problem

Centripetal Force vs Centrifugal Force Wayyyy back in chapter 5, you learned about

something called a “centrifugal force”, which was a fictitious force.

Now that we’ve learned about centripetal forces, we can understand more clearly why it was considered to be fictitious.

Centripetal Force vs Centrifugal Force Imagine that you are on a ride called the

“Merry-go-round twister” – instead of being a classic casual spinning motion, it spins very very fast.

If you are at some point A on the ride, what direction does your velocity vector point?

If there was actually no force acting on you, where would you end up as the ride continued to move?

Centripetal Force vs Centrifugal Force So, what is keeping you on the ride? It is

technically the normal force that the “wall” or “barrier” is pushing you back with – this is was we are calling the centripetal force!

http://www.physicsclassroom.com/mmedia/circmot/cf.cfm

Describing Rotational Motion When an object is constantly rotating, it is

often more useful to describe it by its frequency or period instead of the velocity.

So, we have to find the relationship between the magnitude of the velocity of the object and its frequency and period.

Describing Rotational Motion

Banked Curves Does anyone know why we like to make

“banked”, or tilted, curves instead of flat ones?

An airplane creates a “Lift force” because of the way air travels over the wings – but airplanes can’t change direction like a car: they don’t have wheels to turn!

Banked Curves Cars and trucks can use friction as the

centripetal force, but this may not work if the roads are icy or overly wet.

Instead, roads are made to be “banked” slightly – or on an angle. This way, the normal force in the horizontal (x) direction can create the centripetal force.

Banked CurvesWe can use the following equation to derive a

relationship between the banking angle and the speed of the vehicle rounding the curve.

Banked Curves Notice that the mass of the vehicle does not

affect the amount of banking needed for a vehicle to safely make its way around a curve.

So, a semitrailer and a truck could take the same curve at the same speed as a motorcycle, and wouldn’t have to rely on friction to supply any of the centripetal force.

Banked Curves In the following diagram, a pendulum swings

in a circle. Show that the form of the equation relating the angle (of the string of the pendulum) to the speed of the pendulum bob is identical to the equation for the banking of curves.

Model ProblemCanadian Indy racing car driver Paul Tracy set

the speed record for time trials at the Michigan International Speedway (MIS) in the year 2000. Tracy averaged 378.11km/h in the time trials. The ends of the 3km oval track at MIS are banked at 18.0º and the radius of curvature is 382m.(a) At what speed can the cars round the curves without needing to rely on friction to provide a centripetal force?

(b) Did Tracy rely on friction for some of his required centripetal force?

Model Problem