Physics

Key Unit Competence

By the end of this unit, the learner should be able to analyse and solve problems related to projectile and circular motion.

My goal

By the end of this unit, I will be able to:

* define and explain terms used in projectile motion.

* discuss the different applications of projectile motion.

* apply concepts of projectile and circular motion in real life.

* differentiate between projectile motion and circular motion.

Introduction

We have different kinds of sports, for examples; football, netball, tennis amongst others.

Using the example above

A lot of reasoning is needed while playing football to score one of which is to kick a ball at a certain angle (i.e. to move above the ground). We say that the ball is projected.This also applies to basket ball; the ball to enter the round ring for a score it has to be thrown at a certain angle. Hence, projected.The same principle is used by the military in shooting and launching their missiles.

Projectile Motion

Activity 1: Field study

Aim; To study motion of bodies in free space

a) Out of class, (in pitch,or in school compound), throw a ball,a stone or any body upward.

b) State what happens.

c) Hold a ball in your hands and release it to fall.

d) Is the motion of the ball same as in the first case?

e) Note down your observation.

f) Relate your observations for bodies moving linearly.

Caution

While throwing a stone or any body, take care so that it does not harm you.

We can define a projectile as any body thrown into space/air. The path taken is called a trajectory.

The motion of a projectile unless taken otherwise is a free motion under gravity. We assume that air resistance is negligible in this kind of motion. We have three cases: oblique projection, vertical projection and horizontal projection.

Projection at an angle above the horizontal

• Study the picture below carefully.

• Go outside class and try to kick the ball so that it does not roll on the ground.

• State when will the ball cover a long horizontal range. (State down the conditions for that to occur).

From the figure above, if the ball is kicked so that it does not roll on the ground, it will move at certain angle relative to the ground.

Activity 2

a) In the ground, kick the football individually.

b) By observing, the flight of the ball state whether it will cover a longer horizontal distance when it is projected at a large angle or a small angle.

c) Explain your observation and note down any key points in your book.

Consider a projectile having a certain mass, projected at a speed at an angleα to the horizontal.

Upward projection

From the figure above, a football player can kick the ball and it takes the motion of a projectile.

This is the motion in the x-y plane; we consider axis OX and OY. has two components even the acceleration. We have:

According to OX axis, we have the rectilinear uniform motion whose velocity is constant and has value

v_0x = v₀ cosα = constant

According to OY axis, we have the rectilinear uniformly decelerated motion with acceleration γ₀ sinα – gt

Equations of the Motion

Calculation Of The Maximum Height

The Horizontal Range of the Projectile The horizontal distance travelled by a projectile from the initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range R.

It’s the horizontal distance travelled during the time of flight t_f.

To calculate this range denoted by R, it’s important to know that R = x_max

Velocity at a given point a of the trajectory

Example

1. A particle is projected from a point on a horizontal plane and has an initial velocity of 45 ms-1 at an angle of elevation of . Find the time of flight and the range of the particle on the horizontal plane.

Solution

Vertical Motion

The total time of the flight can be found from the equation:

Downward projection

Note:The motion of the projectile has also two components:

Horizontal projection

Activity 3

Place a stone on top of a table.

Displace it so that its motion takes the shape below.

Try to observe the motion carefully.

Note down what you observe and share it with your class members.

In throwing the stone / displacing it, you should take care so that it does not hit you because it may harm you.

Vertical projection

For this case, we consider the oblique projection where. We find the equations for vertical projection.

Activity 4

Using the information given above;

a) Derive the equations for the motion.

b) Study the picture below and do the same.

c) State the condition when the body attains maximum height.

Circular motion

Activity 5

Study carefully the motion of the ball shown below.State what would happen if at any point the thread holding the ball breaks?Note and record your observation.

The motion is uniform if the body describes in equal angular displacements in equal times..Even if the motion is uniform, it has an acceleration because the velocity changes after every moment since its direction keeps changing.

Angular displacement θ

its equation is: θ =θo +ωt

Where θ: angular displacement of M at time t expressed radian [rad] as S.I of units.

θo = angular displacement of M at t =0 expressed radian [rad] as S.I of units.

ω = angular velocity of M expressed in [rad/s] in S.I of units.

The relationship between S and θ is:

Centripetal acceleration

As we said, in a circular uniform motion, there is acceleration. This acceleration is called centripetal acceleration.

Conclusion: In the uniform circular motion,is carried by the radius and directed toward the centre, reason whyis called the centripetal acceleration.

Periodic time, frequency

Activity 6

* Go to the play ground.

* Make sure you round the playground two times.

* Note and observe the time taken to make one complete revolution.

* What do you call the time taken to move around the play ground.

From the activity 6, you made two rounds in a given time. The number of rounds made in a Unit time is called frequency.

Distance time-graph of a uniform circular motion

When an object executes a circular motion of constant radius R, its projection on an axis executes a motion of amplitude a that repeats itself back and forth, over the same path.

When M executes a uniform circular motion, its projection on X-axis executes a back and forth motion between positions P and P' about O.

Considering the displacement and the time, we find the following graph

Centripetal force

If you try to move / run in a circular path, you will finally notice that you keep moving in a circle even when you try to stop. There is a force that keeps you more in a circular path called centripetal force.

Since a body moving in a circle (or a circular arc) is accelerating, it follows from Newton’s first law of motion that there must be force acting on it to cause the acceleration.

This force, like the acceleration, will also be directed toward the centre and is called the centripetal force. The value F of the centripetal force is given by Newton’s second law, that is:

Wheremis the mass of the body and v is its speed in circular path of radius R. If the angular velocity of the body is ω, we can also say, since v = Rω, F = mRω²

When a ball is attached to a string and is swung round in horizontal circle, the centripetal force which keeps it in a circular orbit arises from the tension in the string.

Other examples of circular motion will be discussed. In all cases, it is important to appreciate that the forces acting on the body must provide a resultant force of magnitude toward the centre.

Application of circular motion

Vertical and horizontal circle

Vertical circle

Taking the approach that the ball moves in a vertical circle and is not undergoing uniform circular motion, the radius is assumed constant, but the speed v changes because of gravity. Nonetheless, the formula of centripetal acceleration is valid at each point along the circle, and we use it at point 1 and 2. The free-body diagram is shown in the figure 8.10 for the positions 1 and 2.

a) At the top (point 1), two forces act on the ball: the force of gravity and the tension force the cord exerts at point 1. Both act downward and their vector sum acts to give the ball its centripetal acceleration. We apply Newton’s second law, for t a vertical direction, choosing downward as positive since the acceleration is downward (toward the centre):

From this equation, we can see that the tension force FT₁ at point 1 will get larger if v₁ (ball’s speed at top of circle) is made larger, as expected.

But we are asked for the minimum speed to keep the ball moving in a circle. The cord will remain taut as long as there is tension in it. But the tension disappears (because v₁ is too small), the cord can go limp and the ball will fall out of its circular path. Thus, the minimum speed will occur if FT₁ = 0, for which we have: (minimum speed at top)We solve for This is the minimum speed at the top of the circle if the ball is to continue moving in a circular path.

b) When the ball is at the bottom of the circle (point 2 in the figure 8.10), the cord exerts its tension force FT₂ upward, whereas the force of gravity, still acts downward. So we apply Newton’s

second law, this time choosing upward as positive since the acceleration is upward (toward the centre):

The second case is the case of a force on revolving ball (horizontal) which is: Estimate the force a person must exert on a string attached to a ball to make the ball revolve in a horizontal circle of radius r.

The forces acting on the ball are the force of gravity, downward, and the tension that the string exerts toward the hand at the centre. The free-body diagram for the ball is as shown in the figure 8.11. The ball’s weight complicates matter and makes it a little difficult to revolve a ball with the cord perfectly horizontal. We assume the weight is small and put θ=0 in the figure 8.11. Thus the tension will act nearly horizontally and, in any case, provides the force necessary to give the ball its centripetal acceleration.

We apply Newton’s second law to the radial direction.

Satellite cycling the earth

The centripetal force which keeps an artificial satellite in orbit round the earth is the gravitational attraction of the earth for it. For a satellite of mass m travelling with speed v in circular orbit of radius (R_E+h) measured from the centre of the earth. R_E is the radius of the earth, h is the height where the satellite is.

Conical pendulum

Activity 7

DO THIS!

* Tie a thread of about 50cm on retort stand.

* On a thread, tie a pendulum bob.

* Displace the bob through a certain angle.

* Displace the bob through a certain angle. What do you observe.

*Release the bob to move through a certain angle so that it moves in a horizontal circle.

* Try to investigate forces acting in the bob.

* Relate your findings to fig. 8.13.

A small object of mass m is suspended from a string of length L. The object revolves with constant speed v in a horizontal circle of radius r, as shown in Figure 8.13. Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.

Let us find an expression for v.

To analyse the problem, begin by letting θ represent the angle between the string and the vertical. In the free-body diagram shown, the force exerted by the string is resolved into a vertical component T cosθ and a horizontal component T sinθ acting toward the centre of revolution. Because the object does not accelerate in the vertical direction, ΣFy = mγy = 0 and the upward vertical component of must balance the downward gravitational force. Therefore, T cosθ = mg (1)

Because the force providing the centripetal acceleration in this example is the component T sinθ, we can use the formula of centripetal acceleration to obtain

A steel ball of mass 0.5kg is suspended from a light inelastic string of length 1 m. The ball swings in a horizontal circle of radius 0.5 m.

Find (i) The centripetal force and tension in the string.

(ii) The angular Speed of the ball.

Road banking

Circular motion on JOB

Activity 8

A car negotiating a corner

The successful negotiation of a bend on a flat road therefore depends on the tyres and the road surface being in a condition that enables them to provide a sufficiently high frictional force, otherwise skidding occurs. Safe cornering that does not rely on friction is achieved by “banking” the road.

The problem is to find the angle αat which the bend should be banked so that the centripetal force acting on the car arises entirely from a component of the normal forceof the road.

Treating a car as a particle and resolving vertically and horizontally we have; since N sinα is the centripetal force wheremand v are the mass and the speed respectively of the car and R is the radius of the bend. Also, the car is assumed to remain in the same horizontal plane and so has no vertical acceleration, therefore N cos α = mg.

The equation shows that for a given radius of bend, the angle of banking is only correct for one speed.

Career centre

Learn more about careers in physics where projectile and circular motion are applied.

Exercises

Projectile motion

1. A projectile is thrown horizontally with a speed of 300m/s from the top of a building 78.4m high.

a) Compute the range of the projectile.

b) What is the time taken to reach the ground?

2. A machine gun throws a projectile is  with a speed of 740m/s Find the range, the maximum height reached by the projectile and the time taken to reach the ground, when it is projected through an angle of 45°.

3. The range of the motion of a projectile is m when it is projected from the ground with a speed of 20m/s What is the maximum height reached? g = 10m/s².

4. A projectile thrown through an angle of 30° reaches 50m. Find the initial speed.

5. A football player punts the ball so that it will have a “hang time” (time of flight) of 4.5s and land 45.7m away. If the ball leaves the player’s foot 1.52m above the ground, what is its initial velocity (magnitude and direction)?

6. A projectile is fired with an initial speed of 400m/s at an angle of 60° above the horizontal from the top of a cliff 49m high. Determine the:

a) time to reach the maximum height.

b) maximum height above the base of the cliff reached by the projectile.

c) total time it is in the air.d) horizontal range of the projectile.

7. A projectile is fired with a speed of 600m/s at an angle of 60°. Find:

a) the horizontal range.

b) the maximum height.

c) the speed and the height after 30s.

d) the time and the speed when the projectile reaches 10km.

e) the time to reach the maximum height.

8. From the top A of a cliff 100m high, a projectile is fired at angle 45°. The initial speed is 400m/s and g = 10m/s². Determine:

a) The maximum height.

b) The horizontal distance below A and the point where it strikes the ground.

c) The time taken to travel the total distance.

d) The velocity when it strikes the ground.

9. A rescue plane is flying at a constant elevation of 1200m with a speed of 430km/h toward a point directly over a person struggling in the water. At what angle of sight θ should the pilot release a rescue capsule if it is to strike (very close to) the person in the water?

10. A police officer is chasing a burglar across a roof top; both are running at 4.5m/s Before the burglar reaches the edge of the roof, he has to decide whether or not to try jumping of the next building, which is 6.2m away but 4.8m lower. Can he make it? Assume that he jumps horizontally.

Extension questions

1. A bomber (plane) is at an altitude of 20km and has 400km/s of speed. When the plane is above a certain point, it releases a bomb. How long will it take the bomb and at what distance from the same point to reach the ground?

2. A gun fires a projectile of mass 2kg at initial velocity V₀ = 200m/s making an angle θo = 53° with the ground.

a) Find;

(i) The kinetic energy of the projectile at the point where it leaves the gun.

(ii) The potential energy at the highest point of the flight; deduce the velocity of the projectile at that point.

(iii) The range of the projectile; what is the time taken for the flight?

b) Find the position of the projectile, the magnitude and direction of the velocity when t = 25.3. A ball is thrown horizontally with a speed V₀ of 243.8cm/s. Find the position and velocity after 25.

4. A ball is thrown vertically upward in air and returns in the hand from which it was 3s later. A second ball is thrown at an angle of 30o with the horizontal. At which velocity must the second ball be thrown so that it reaches the same maximum height as the first one thrown vertically?

5. a) A projectile is launched with speed at angle αo above the horizontal. The launch point is at a height h above the ground. Show that if air resistance is ignored, the horizontal distance that the projectile travels before striking the ground is 

b) Determine x if h is taken to be zero.

Circular motion

1. A body describes a circumference of radius 2m and the motion is uniform. It does 2 rotations in 6s. If π² = 9.86, find the centripetal acceleration.

2. A moving body is in uniform circular motion. The radius of the circle is 25m. Assuming that the acceleration equals 9m/s², find the angular velocity.

3. How many rotations a wheel of 3.20m diameter does in one minute. Assuming that the linear speed is equal to 16m/s?

4. A ball at the end of a string is swinging in a horizontal circle of radius 1.15m. The ball makes exactly 2.00 revolutions in a second. What is its centripetal acceleration?

5. The wheel of an engine of 4m diameter does 90 rotations per minute. Calculate:

a) the linear speed.

b) the angular speed.

c) the centripetal acceleration.

6. What is the angular velocity of the earth around its axis? What is the linear velocity of a point situated at the equator? The radius of the earth is supposed to be 6400km.

7. The motion of the earth is considered as a rotation about its axis. Find, for a point located in Paris (latitude 45°),

a) The angular velocity.

b) Linear velocity.

c) Centripetal acceleration.

8. Two bodies A and B describe in the same direction, the same circle of radius 10cm with constant angular velocities WA = 10 rad/s and WB = 11 rad/s. The motion starts when they are at the origin.

a) After how many seconds do they coincide again for the first time?

b) What is the distance travelled by A?

9. The artificial satellite syncom appears motionless in the sky and its trajectory is circular at the height of 35,700km. What is its speed? The radius of the earth is 6400km.

10. What time is it, after 12:00 the watch hands make an angle of 180° for the first time?

11. How many minutes after 4:00, watch hands coincide for the first time?

12. What time is it, after 3:00, the watch hands make a right angle for the first time?

13. An engine having a speed of 4000 rotations per minute decelerates during 8s till the stop. How many rotations does it make in that time?

14. Let a uniformly decelerated circular motion of deceleration 0.5 rads². At t = 1 sec, the angular velocity has the magnitude of 1rad/s and the body is at the point . Find the equation of the motion.

15. Let θ = 5+ 4t −t² be the equation of a circular motion of radius 0.05m.

a) determine the angular velocity at t = 0 and t = 2.

b) determine its angular acceleration.

c) determine its acceleration and the linear velocity at t = 0 and t = 2s.

16. A punctual moving object describes a circular trajectory of radius r= 18m. The curvilinear displacement is given by S=3t², where S is in meter and t in seconds. Calculate the acceleration at t = 2s.

17. During 5s a wheel doubles its angular velocity and executes 120 rotations. What are the magnitudes of the angular velocities at the beginning and the end of the process?

Extension questions

1. An electric motor is switched off and its angular velocity decreases uniformly from 900 rotations to 400 rotations in 5s.

a) Find the angular acceleration in rotations/sec² and the number of rotations done by the motor in the time interval of 5s.

b) How long does it take the motor to stop if the angular acceleration remains constant and equal to the one in (a)?

Dynaics of circular motion, centripetal force

1. Compute the centripetal force applied on a wheel of mass 1000kg, assuming that its diametre is 3m and it turns with a speed of 300 rotations per minute.

2. A train of 105kg travels with a speed of 70km/hand reaches a bend of radius 500m. Find the value of the centripetal force.

3. A frigate bird is soaring in a circular path. Its bank angle is estimated to be 25° and it takes 13s for the bird to complete one circle.

4. How fast is the bird flying?

5. What is the radius of the circle?

6. A 1000kg car rounds a curve on a flat road of radius 50m at a speed of 50 km/h. Will the car make the turn if:

(a) The pavement is dry and the coefficient of static friction is 0.60,

(b) The pavement is icy and the coefficient is 0.20?Calculate the speed required for a satellite moving in a circular orbit 200km above the earth’s surface.

7. What is the maximum speed with which a 1300kg car can round a turn of radius 95m on a flat road if the coefficient of friction between tires and road is 0.55? Is this result independent of the mass of the car?

8. How large must the coefficient of friction be between the tires and the road if a car is to round a level curve of radius 62m at a speed of 55km/h?

9. If a curve of radius of 60m is properly banked for a car traveling 60km/h, what must be the coefficient of static friction for a car not to skid when traveling at 90km/h?

10. A 1200kg car rounds a curve of radius 65m banked at an angle of 14°. If the car is traveling at 80km/h, will a friction force be required? If so, how much and in which direction?

11. A vehicle of mass 1000kg is moving on a bridge which has the shape shown on the figure below. The radius is of 50m and the speed is 15m/s Find the magnitude of the force exerted by the vehicle on the bridge if the car is on the top of the bridge.

12. A mass m₁ on a frictionless table is attached to a hanging mass m₂ by a cord through a hole in a table (see the figure). Find the speed with which m₁ must move for m₂ to stay at rest.

Extension questions

1. A mass rotates on a vertical circle at the end of a light string of 0.3m of length. Calculate:

a) The difference in kinetic energy between the upper and the lower point of the circle.

b) The difference in tension of the string between the upper and the lower point of the circle.

2. Read the passage below and answer the questions that follow: “Satellites which orbit 35,788km above the Earth’s equator are said to be in geostationary orbit. Most telecommunication satellites are in this type of orbit. Other satellites are in solar orbit only and few hundreds of miles above the earth. These are said to be in low earth orbit. These are mainly military satellites which because of their low orbit can “see” things in great detail”,

What is the difference between an artificial satellite and a natural satellite of the Earth?

c) Briefly explain three benefits of artificial satellites.

d) Describe what each of the following words used in the passage mean:

(i) Geostationary,

(ii) Low earth orbit.

e) Establish an expression for the velocity of geostationary satellites in terms of altitude h, earth radius R and acceleration due to gravity at the earth’s surface g and then find its numerical value.

3. a) An object is moving in a circle with constant speed. What is the direction of the net force acting on this object?

b) What is the net force required to make an object of 40kg accelerate at a rate of 2m/s?