Grade 12 Physics Past Paper - Two-Dimensional Motion

Unit 2 Brain storming question 2.1 Two-dimensional motion 1. Consider a ball Introduction shot horizontally from a very high Kinematics is the study of motion without considering its causes. For example, building at a high studying the motion...

Unit 2 Brain storming question 2.1 Two-dimensional motion 1. Consider a ball Introduction shot horizontally from a very high Kinematics is the study of motion without considering its causes. For example, building at a high studying the motion of a football without considering what forces cause or speed. Assume that change its motion. Two-dimensional kinematics are simple extensions of the there is no force of one-dimensional kinematics developed for motion in a straight line in Grade 11. gravity acting on the ball. What would This simple extension will allow us to apply physics to many more situations, and the motion of the it will also yield unexpected insights about nature. ball be like? Explain A ball kicked by a football player, the orbital motion of planets, a bicycle its motion? rounding a curve,the rotation of wheels of a car are a few examples of 2. The ball is two-dimension motion. In fact, most motions in nature follow curved paths projected rather than straight lines. Such types of motion along a curved plane are horizontally from described by two-dimensional kinematics. the top of the same building. This time, At the end of this unit, you will be able to: the force of gravity Understand the basic ideas of two-dimensional motions. is acting on the ball. What will the Describe the motion of objects in horizontal and inclined projectiles; motion of the ball be like? Will gravity Describe uniform rotational motion,rotational dynamics and Kepler’s affect the ball’s laws horizontal motion? Will the ball travel a Describe Newton’s law of Universal gravitation. greater (or shorter) horizontal distance Develop pertinent problem-solving skills. due to the influence of gravity? 24 2.1 Projectile motion 25 2.1 Projectile motion At the end of this section, you will be able to: Explain the motion of the projectile with respect to the horizontal and vertical components of its motion. Derive equations related to projectile motion. Discussion Apply equations to solve problems related to projectile motion question 2.1 A projectile is a thrown, fired, or released object that moves only under the Which motion is different from the influence of gravitational force. The projectile accleration is g = 9.8m/s s. Anyone others? Explain who has observed the motion of a ball kicked by a football player (Figure 2.1b) Why? has observed projectile motion. The ball moves in a curved path and returns to a) A ball thrown the ground. Other examples of projectile motion include a cannonball fired from horizontally into the a cannon, a bullet fired from a gun, the flight of a golf ball and a jet of water air. escaping a hose. b) A bullet fired from a gun. c) A javelin thrown by an athlete. d) A bird flying in the air. Figure 2.1 a) A ball thrown horizontally b) A football kicked in a game Projectile motion of an object is simple to analyze if we make three assumptions: 1. The free-fall acceleration is constant over the range of motion, and it is always directed downward. It is the acceleration due to gravity (g)=9.8m/s 2. 2. The effect of air resistance is negligible. 3. The horizontal velocity is constant because the acceleration of the object does not have vertical component. With these assumptions, we find that the path of a projectile, which we call its trajectory, is a parabola as shown in Figure ??. The horizontal and vertical components of a projectile’s motion are completely independent of each other and can be handled separately, with time t as a 26 Unit 2 Two-dimensional motion common variable for both components. Horizontal Projection In this type of motion the projectile is projected horizontally from a certain height as shown in Figure 2.2. Its initial velocity along the vertical direction is zero and it possesses only horizontal velocity at the beginning. As the time progresses, due to the impact of gravity, it acquires the vertical component of velocity (Figure 2.2). Equations for the horizontal component of motion Figure 2.2 The motion of a ball The projectile has zero acceleration along x direction. Therefore, the initial projected horizontally. velocity v0x remains constant throughout the motion. We use constant acceleration motion equations. The final horizontal velocity, v x after a time t is: vx = v0x (constant) The horizontal distance traveled by the projectile at a time t is given by the equation Discussion question 2.2 ∆x = v 0x t (2.1) Assume that an airplane flying Equations of vertical motion horizontally drops a package to a remote The vertical motion is a constant accelerated motion. We use the kinematic village. equations of motion for constant accelerated motion. The final vertical velocity What kind of motion v y after time t is: is performed by the package? v y = v oy + g t (2.2) Draw the trajectory of the package. Where v0y is the initial vertical velocity. As the package hits the ground at the The initial vertical velocity has no downward component (v0y =0). Therefore village, where is the aircraft? v y = gt From the kinematics equations, the vertical displacement, ∆y has a form: 1 ∆y = v 0y t + g t 2 (2.3) 2 But v0y = 0, therefore 2.1 Projectile motion 27 ∆y = 21 gt2 Remember: When you use equations to answer questions on vertical motion, upwards motion is positive (+) and downwards motion is negative (-). Time of flight Activity 2.1 The time of flight is the time taken by the projectile to hit the ground. Place two tennis We know that: balls at the edge of a tabletop. Sharply ∆y = 21 gt2 snap one ball horizontally off the Then table with one hand while gently tapping s 2∆y t= the second ball off g with your other hand. Measure Range the height (y) of The range is the maximum horizontal distance traveled by the projectile. the table and the Once we find the time of flight t, we can solve for the horizontal displacement horizontal distance between the table’s using: edge and the balls ∆x = v0x t landing location (R). Determine the In projectile motion, the time to cover both the x and y displacement is the same. following from your By substituting the total time flight, we get: measurements: a) The time of flight s 2∆y of both tennis R = v0x g balls. Explain your result. b) The initial Example 2.1 horizontal velocity A rifle is aimed horizontally at a target 30m away as shown in Figure 2.3. The bullet of the balls when hits the target 2 cm below the aiming point. they leave the (a) What is the bullet’s time of flight? edge of the table. (b) What is the initial velocity of the bullet? Assume gravity (g) =10m/s 2. Solution: The givens in this question are: ∆X =30 m, ∆Y =2 cm =0.02 m, g=10m/s 2. (a) The equation for the vertical displacement is: 28 Unit 2 Two-dimensional motion 1 2 ∆y = gt 2 1 -0.02 m = (-10)t2 2 The vertical displacement is in the negative direction, which gives: t=0.06 s Since this is the time of impact with the target, the time of flight of the bullet is also Figure 2.3 A bullet fired the same. horizontally. (b) The equation for x−motion is: ∆x =vox t ∆x 30m vox = = t 0.06s The initial velocity of the bullet is 500 m/s. Example 2.2 A rescue airplane travelling at 360 km/h horizontally dropps a food package from a height of 300 m when it passes over a car driver stranded in the desert. Assumming (g)=10m/s 2. (a) How long will it take the food package to reach the ground? (b) How far from the car driver should the food package be dropped ? Solution: (a) The package has the same horizontal velocity as the airplane. Therefore, the initial vertical velocity is zero. The equation for the vertical displacement is: 1 2 ∆y = gt 2 1 -300 m = (-10)t2 2 The vertical displacement is in the negative direction, which gives: t =7.74 s (b) The equation for the horizontal displacement is: 2.1 Projectile motion 29 ∆x =v0x t ∆x =100 m/s x 7.74 s ∆x =774 m Activity 2.2 Use this activity to investigate horizontal projection. Materials Ruler A cannon ball made from scrunched up aluminum foil. Rubber band. A tube made from paper or cardboard with diameter larger than the diameter of the ball. Procedures: 1. Put the tube near the edge of the table. 2. Use the rubber band to shoot the ball out of the tube. 3. Stretch the rubber band the same amount each time to make sure the initial velocity is constant. 4. You can increase the stretching of the rubber band to increase the initial horizontal velocity of the projectile. 5. Measure the height of the table. Use this height to calculate the time of flight (assume there is no air resistance). 6. Measure the horizontal distance traveled by the canon ball. Use this distance to calculate the initial velocity of the projectile. Inclined projectile motion This is a type of motion in which an object is projected with an initial velocity v0 which makes an angle θ with the horizontal (Figure ??).The initial velocity can be resolved into two components, vertical and horizontal component. The vertical component of the velocity changes with time as a result there is acceleration due 30 Unit 2 Two-dimensional motion to gravity. The horizontal component of the velocity is constant throughout the flight; this is because there is no force acting along the horizontal direction of the projectile as a result there is no acceleration along x-axis. The analysis of the motion involves dealing with the two motions. As shown in Figure 2.4, the projectile has velocity components at different positions. At the top where it reaches its maximum height the vertical Discussion component of the velocity becomes zero. After V y becomes zero the projectile question 2.3 changes its direction and make free fall. Balls A and B are kicked at an angle of 370 and 530 with the horizontal respectively, with the same initial velocity v0. Which ball has: a) the maximum horizontal displacement? b) the maximum height? Figure 2.4 Inclined projectile motion. Equations of inclined projectile motion The initial velocity can be expressed as x component and y component: v0x = v0 cos θ v0y = v0 sin θ The horizontal velocity at any time t is: vx = v0 cos θ (constant) The vertical velocity at any time t is: v y = v o si n θ + g t (2.4) 2.1 Projectile motion 31 Displacements of the projectile There are two different types of displacement of the projectile motion: Horizontal displacement at any time t: ∆x = v 0 cosθt (2.5) Vertical displacement at any time t: 1 ∆y = v 0 si nθt + g t 2 (2.6) 2 The time to reach the maximum height is: v y = v0 sinθ + gt Since v y = 0 at maximum height and g is negative: v 0 sin θ t= g Time of flight The time of flight is the total time for which the projectile remains in flight. The time of flight depends on the initial velocity of the object and the angle of the projection, θ. 1 ∆y = v0 sin θ t + gt 2 2 When the point of projection and point of return are on the same horizontal level, the net vertical displacement of the object is zero, ∆Y =0. 1 0 = v0 sin θ t + gt 2 2 Apply factorization, we have: 0 = t(v0 sin θ + 21 gt) Since t cannot be zero and g is negative,solving for t gives us: 2v 0 sin θ tt ot al = g This last equation does not apply when the projectile lands at a different elevation from the one at which it was launched. 32 Unit 2 Two-dimensional motion Horizontal range and maximum height of a Projectile Let us now consider a special case of projectile motion. Assume a projectile is launched from the origin at O, as shown in Figure 2.4, and returns to the same horizontal level. This situation is common in sports, where baseballs, footballs and golf balls often land at the same level from which they were launched. Two points in this motion are especially interesting to analyze: the peak point A, which has Cartesian coordinates (R/2, H), and the point B, which has coordinates (R, 0). The distance R is called the horizontal range of the projectile, and the distance H is its maximum height. Let us find R and H mathematically in terms of v0 , θ, and g. Range(R) The range of the projectile is the maximum displacement in the horizontal direction. There is no acceleration in this direction since gravity only acts vertically. ∆x = v0 cos θ t When ∆x is maximum, ∆x=R. Since the time to cover the range is the total time of flight: 2v 0 sin θ tt ot al = g R = v0 cos θ tt ot al v 0 2 sin 2θ R= g This equation is valid for launch and impact on a horizontal surface, as shown in Figure 2.5. We can see in Figure 2.5a the range is directly proportional to the square of the initial speed v0 and sin2θ. Furthermore, we can see from the factor sin2θ that the range is maximum at 45◦. In Figure 2.5 (a) we can see that the greater the initial velocity, the greater the range. In Figure (b) the range is maximum at 45◦. This is true only for conditions ignoring air resistance. It is interesting that the same range is found for two initial launch angles that add up to 90◦. The projectile launched with the smaller angle has a lower peak than the higher angle, but they both have the same range. 2.1 Projectile motion 33 Figure 2.5 Trajectories of projectiles on leveled ground. (a) The effect of initial velocity v0 on the range of a projectile with a given initial angle. (b) The effect of initial angle θ on the range of a projectile with a given initial speed. Maximum height (H) The maximum height of a projectile trajectory occurs when the vertical component of velocity, v y equals zero. As the projectile moves upwards it goes against gravity, and therefore the velocity begins to decrease. Eventually the vertical velocity will reach zero, and the projectile is immediately accelerated downward under gravity. Thus, once the projectile reaches its maximum height, it begins to accelerate downward. 1 ∆y = v0 sin θ t + gt 2 2 v 0 sin θ The time to cover the maximum height is: t = g When ∆y is maximum, ∆y =H v 0 2 sin2 θ H= 2g Discussion question 2.4 1. A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection? 2. A ball is kicked into the air from the ground at an angle θ with the horizontal. When the ball reaches its highest point, which statement is true ? (a) Both the velocity and acceleration of the ball are zero. 34 Unit 2 Two-dimensional motion (b) Its velocity is not zero, but its acceleration is zero. (c) Its velocity is perpendicular to its acceleration. (d) Its acceleration depends on the angle at which the ball was thrown. Activity 2.3 3. One ball is thrown horizontally. At the same time, a second ball is Use this activity to dropped from the same height. Ignoring air resistance and assuming the investigate inclined ground is level, which ball hits the ground first? Explain why. projection. You need the materials listed Relation between range and maximum height in Activity 2.2. Procedures: Consider a projectile motion as shown in Figure 2.4. The initial velocity of the (a) Adjust the tube projectile is v0 , H is the maximum height and R is its horizontal range. We know at different that the maximum height of the projectile H is given by the equation: angles from the horizontal. (b) As before keep v 0 2 sin2 θ H= the stretching of 2g the rubber band And horizontal range is given by the equation: constant. (c) Vary the angle of v 0 2 sin 2θ projection. R= g (d) Measure the Divide the maximum height of the projectile by the horizontal range. relationship (In the equation, sin2 θ can be written as sinθsinθ, and sin2θ can be written between the angle as 2sinθcosθ). of projection, H si nθ range and = R 4cosθ maximum height R t anθ reached by the H= 4 projectile. Example 2.3 A football player kicks a ball at angle of 370 with the horizontal. The initial velocity of the ball is 40 m/s. a) Find the maximum height reached by the ball. b) Find the horizontal range of the ball. 2.1 Projectile motion 35 Solution: In this problem the given quantities are: v0 = 40 m/s, θ = 370 and g=10 m/s2 a) The maximum height reached is: V0 2 sin2 θ H= 2g (40m/s)2 sin 370 sin 370 H= 2x10m/s 2 H=28.8 m b) The horizontal range is: V0 2 sin 2θ R= g (40m/s)2 sin 740 R= 10m/s R=153.8 m Example 2.4 A ball is kicked from the ground with an initial speed of 25 m/s at an angle of 530 above the horizontal directly toward a wall, as shown in Figure 2.6. The wall is 24 m from the release point of the ball. (a) How long does the ball take to reach the wall? (b) How far above the ground level does the ball hit the wall? (c) What are the horizontal and vertical components of its velocity as it hits the wall? (d) What is the resultant velocity with it hits the wall? Solution: In this problem the given quantities are: ∆x = 24 m, θ =530 , v0 =25 m/s (a) The horizontal displacement of the ball is given by the equation ∆x = v0 cosθt Solving for the time at which ∆x = 24 m: ∆x 24m t= = v 0 cosθ (25m/s)x0.6 t = 1.6 s Thus, the ball reaches the wall 1.6s after being thrown. 36 Unit 2 Two-dimensional motion (b) We can answer this question if we can find the y coordinate of the ball at the time it hits the wall, namely at t = 1.6 s. We need the y equation of motion. 1 ∆y = v0 sin θ t + gt 2 2 1 ∆y = 25 m/s x0.8 x 1.6 s + x(−10m/s 2 )x(1.6s)2 2 ∆y =19.2 m This tells us that the ball hits the wall at 19.2 m above the ground level. (c) The x and y components of the ball’s velocity at the time of impact (t=1.6 s) vx =v0 cosθ vx = 25 m/s x 0.6 vx =15 m/s v y =v0 sinθ +gt Figure 2.6 A ball thrown toward a v y = 25m/s x 0.8 +(−10m/s 2 )x1.6 s wall. v y =4 m/s (d) The resultant velocity is the vector sum of the x and y components. q v= v x2 + v 2y p v= (15m/s)2 + (4m/s)2 v= 15.5 m/s Activity 2.4: Use this activity to investigate inclined projection. In this activity you use the law of conservation of mechanical energy that you learned in grade 11. Materials required Ruler V-shaped track with a shorter launch track. Small ball (e.g. tennis ball). protractor Procedures: 1. Adjust the shorter end of the track to the edge of the surface of a table. 2. Use a short segment of the track at an angle; say 45 degrees with respect to the surface of a table. 2.1 Projectile motion 37 3. Measure the height of the longer end of the track where the ball is to be released and also measure the height of the shorter end where the ball is going to leave the surface of the table. 4. Put the ball in motion down the track. 5. Calculate the speed of the ball on the track just as it leaves the level of 1 the surface of the table using conservation of energy (mgh = mv2 ). 2 6. Calculate the time it takes to fall back to the surface of the table 2V0 sin θ t=. g V0 2 sin 2θ 7. Predict where the ball will land using, x=. g 8. Put a cup there to catch the ball. Put the ball in motion down the track again. 9. Change the velocity of the ball by changing the inclined angle of the longer arm of the track (this is to reduce the height from which the ball is released). Discussion question 2.5 1. As a projectile moves in its parabolic path, is there any point along the path where the velocity and acceleration vectors are (a) perpendicular to each other (at right angles)? (b) parallel to each other? 2. Which of the following statements about projectile motion are true? (ignoring air resistance). (a) The horizontal and vertical motions are independent. (b) The force on the projectile is constant throughout the flight. (c) The acceleration of the projectile is constant throughout the flight. (d) The path depends upon the initial velocity, but not upon the mass of the projectile. (e) All of the above statements are true. 3. A projectile is fired on Earth with some initial velocity. Another projectile 38 Unit 2 Two-dimensional motion is fired from the surface of the Moon with the same initial velocity. If air resistance is ignored, which projectile has the greater range? Why? Which reaches the greater height? Why? (Note that the free-fall acceleration on the Moon is about 1.6 m/s2 ). Exercise 2.1 Use g=10m/s 2 where necessary. 1. At which position in its flight will a ball experience its minimum speed during inclined projection? A. at the beginning B. at maximum height C. at the end D. the same speed at all positions 2. A gun with a muzzle velocity of 500 m/s shoots a bullet at a target 50 m away. To hit the target the gun should be aimed: A.directly towards the target along the line joining the gun and target. B.10 cm high above the target. C. 5 cm high above the target. D. 5cm below the target. 3. A ball is thrown horizontally with a velocity of 20m/s from a top of building 90 m high. Calculate: a) the time taken to reach the ground. b) the horizontal displacement. c) The resultant velocity with which it strikes the ground. 4. A long jumper leaves the ground at an angle of 20.0o above the horizontal and at a speed of 11.0 m/s. a) How far does he jump in the horizontal direction? b) What is the maximum height reached? 5. An object projected at an angle θ with velocity 30 m/s reaches its maximum height in 1.5 s. Calculate its range. 2.2 Rotational Motion 39 2.2 Rotational Motion At the end of this section, you will be able to: Describe the motion of a rigid object around a fixed axis. Derive equations of motion with constant angular acceleration. Apply equations to solve problems related to rotational motion. Rotational motion is the motion of an object in a circle around a fixed axis. For example, the rotation of Earth around its axis, the rotation of the flywheel of a sewing machine, rotation of a ceiling fan, rotation of wheels of a car, and so on. Figure 2.7 Rotation of a disc of mass M around a fixed axis. The disc in Figure 2.7 is performing rotational motion because all of its particles are rotating around a fixed axis, called its axis of rotation. An object can rotate around a fixed point in two directions: a clockwise or an anticlockwise direction (also known as counterclockwise). Rigid body is n object with a perfectly defined and unchanging shape. NO matter the size of the force , the distance between any two particles within the object remains constant. Angular displacement and angular velocity Angular displacement(∆θ) Figure 2.8 is a view from above of a rotating compact disc, or CD. The disc rotates Figure 2.8 A CD rotating about a around a fixed axis perpendicular to the plane of the figure, passing through the fixed axis through O perpendicular center of the disc at O. One particle of the disc P, is kept at a fixed distance r from to the plane of the figure. the origin and rotates around O in a circle of radius r. Because the disc is a rigid object, as the particle moves through an angle θ from the reference line, every other particle on the object rotates through the same angle. Therefore, we can associate the angle θ with the entire rigid object as well as with an individual particle. Figure 2.9 A particle P on a rotating disc moves from A to B along the arc of a circle. As the particle travels from position A to position B in a time interval ∆t ,as shown in Figure 2.9, the line joining the particle to the center sweeps out an angle ∆θ. This quantity ∆θ is defined as the angular displacement of the rigid object. 40 Unit 2 Two-dimensional motion ∆θ = θ f − θ 0 (2.7) Because rotational motion involves studying circular paths, rather than using meters to describe the angular displacement of an object, physicists use radians or degrees. A radian is convenient because it naturally expresses angles in terms of π since one complete turn of a circle (360 degrees) equals 2π radians. 1revolution = 2πrad = 3600 Angular velocity(ω) How fast an object is rotating can be calculated using the concept of angular velocity. If the disc spins rapidly, the angular displacement can occur in a short time interval. If it rotates slowly, the angular displacement occurs in a longer time interval. The rate at which angular displacement occurs can vary. These different rotation rates can be quantified by defining the average angular velocity ωav (Greek letter omega) as the ratio of the angular displacement of a disc to the time interval ∆t during which the displacement occurs. θ f − θ0 ∆θ ωav = = (2.8) t f − t0 ∆t Angular velocity has units of radians per second (rad/s). Angular acceleration If the angular velocity of an object changes from ω0 to ω f in the time interval ∆t, the object has an angular acceleration. The angular acceleration α (Greek letter alpha) of a rotating rigid object is defined as the ratio of the change in the angular speed to the time interval ∆t during which the change in the angular speed occurs: ω f − ω0 ∆ω α= = (2.9) t f − t0 ∆t Angular acceleration has units of radians per second squared (rad/s2 ). 2.2 Rotational Motion 41 Direction of angular velocity and angular acceleration Angular velocity and angular accelerations can be treated as a vectors, so we Discussion must include magnitude and direction. For rotation around a fixed axis, the question 2.6 direction of rotational motion is specified in relation to the direction along the In small group, axis of rotation. Therefore, the directions of ω and α are along this axis. discuss how to find To illustrate this convention, it is convenient to use the right-hand rule the direction of demonstrated in Figure 2.10. When the four fingers of the right hand are wrapped angular velocity in the direction of rotation, the extended right thumb points in the direction of ω. and angular The direction of α follows from its definition α = ∆ω /∆t. It is in the same acceleration. direction as ω if the angular speed is increasing in time, and it is antiparallel (parallel but moving in the opposite direction) to ω if the angular speed is decreasing in time. Figure 2.10 The right-hand rule for determining the direction of the angular velocity vector. Equation of motion for constant angular acceleration Consider a rigid object such as the CD rotating around a fixed axis with a constant angular acceleration. A set of kinematic equations exist for rotational motion just as they do for translational motion. They have a similar form and are derived in a similar fashion. ωf − ω0 α= (constant angular acceleration) (2.10) tf − t0 Then, by rearranging, we get an equation 42 Unit 2 Two-dimensional motion ω f = ωo + α∆t (2.11) where ωo is the angular speed of the rigid object at time t = 0. This equation allows us to find the angular speed ω f of the object at any later time t. If the angular acceleration is constant, the average angular velocity is obtained by: ωo + ω f ωav = 2 ∆θ ωav = ∆t Combining these two equations, you we get: ∆θ ωo + ω f = ∆t 2 When we substitute ω f = ωo + α∆t 1 ∆θ = ωo ∆t + α∆t 2 (2.12) 2 This equation allows us to find the angular displacement of the object at any later time t. ωo + ω f ω f − ωo µ ¶ We know that: ∆θ = ∆t and ∆t = 2 α Combining these two equations, we get: ωo + ω f ω f − ωo µ ¶µ ¶ ∆θ = 2 α ω2f = ω20 + 2α∆θ (2.13) This equation allows us to find the angular speed ω f of the rigid object for any value of its angular position ∆θ. Example 2.5 What is the average angular velocity of a rotating wheel if its angular speed changes from 30 rad/s to 50 rad/s in 2 s? 2.2 Rotational Motion 43 Solution: In this problem the given quantities are: ωi =30 rad/sec, ω f = 50 rad/s and t=2 s ω f − ω0 αav = ∆t 50r ad /s − 30r ad /s αav = 2s αav = 10 r ad /s 2 Example 2.6 A rotating wheel has an initial angular velocity of 10 rad/s and accelerates at 2.5 rad/s2. (a) How many revolutions are completed in 30 s? (b) What is angular speed of the wheel at t =20 s? Solution: The given quantities are ωo = 10 rad/s, α = 2.5 rad/s2 , t=30 s (a) ∆θ = ωo ∆t + 21 α∆t 2 Substitute the known values to find the angular displacement. 1 ∆θ=10rad/sx30s + x2.5r ad /s 2 x900s 2 2 ∆θ = 300 rad + 1125 rad ∆θ = 1425 rad Convert rad in to revolution 1rev = 2π rad µ 1r ev ¶ ∆θ = 1425 rad 2πr ad ∆θ = 226.9 rev (b) The final speed at t = 20 s is asked ω f = ωo + α∆t ω f = 10 rad/s + 2.5 rad/s2 x20 s ω f = 60 rad/s Example 2.7 A car’s wheel has an initial angular velocity of 6 rad/s and a constant angular acceleration of 3 rad/s2. Calculate the angular velocity after 100 rev ? 44 Unit 2 Two-dimensional motion Solution: The given quantities are: ωo = 6 rad/s, α = 3 rad/s2 , ∆θ = 100 rev First convert rev to rad. 1rev = 2πrad ³ ´ 2πr ad ∆θ = 100 rev 1r ev ∆θ = 628 rad ω2f = ω20 + 2α∆θ ω2f =(6r ad /s)2 + 2x(3r ad /s 2 )2 x628r ad ω f = 61.68 rad/s Kinematic equations for rotational and linear motion The kinematics for rotational motion is completely analogous to linear (or translational) kinematics. Many of the equations for the mechanics of rotating objects are similar to the motion equations for linear motion. When solving problems involving rotational motion, we use variables that are similar to linear variables (distance, velocity and acceleration) but take into account the curvature or rotation of the motion. We defined: the angular rotation ∆θ, which is the angular equivalence of distance, ∆s; the angular velocity ω, which is the angular equivalence of linear velocity v; the angular acceleration α, which is the angular equivalence of linear acceleration, a. Example 2.8 A wheel has a radius of 20 cm and accelerates from rest to 15 rev/s in 30 s. What is the magnitude of the tangential acceleration of a point at the tip of the wheel. Solution: The angular acceleration is: ω f − ωo 15r ev/s − 0 α= = ∆t 30s α =0.5 r ev/s 2 2.2 Rotational Motion 45 Table 2.1 shows the analogy between linear and angular motion equations. Linear motion with Angular motion with constant acceleration constant angular acceleration v f = v0 + a∆t ω f = ωo + α∆t v f + v0 ω f + ω0 Vav = ωav = µ 2 2 ¶ v f + v0 ω f + ω0 ¶ µ ∆s = ∆t ∆θ = ∆t 2 2 1 1 ∆s = v0 t + a∆t 2 ∆θ = ωo ∆t + α∆t 2 2 2 v f 2 = v0 2 + 2a∆s ω f 2 = ω20 + 2α∆θ Since 1rev = 2π rad α =3.14 r ad /s 2 Therefore, the tangential acceleration is at = αr at = 3.14 rad/s2 x 0.2 m at = 0.6 m/s2 Example 2.9 A car accelerates from 20 m/s to 24 m/s in 5 s. Calculate the angular acceleration of the wheels of the car if the radius of a wheel is 40 cm. Solution: First, we calculate the tangential acceleration of a point on the rim of the wheel. The equation to use is: v f − v o 24m/ sec −20m/s at = = ∆t 5s a t = 0.8 m/s 2 Then the angular acceleration of the wheels is: at α= r 0.8m/s 2 α= 0.4m α = 2 r ad /s 2 46 Unit 2 Two-dimensional motion Example 2.10 A boy rides a bicycle for 5 minutes. The wheel with radius of 30 cm completes 2000 rev during this time. Calculate. (a) the average angular velocity of the wheel. (b) the linear distance traveled by the bicycle in 5 minutes. Solution: r = 30 cm = 0.3 m, ∆θ = 2000 rev ∆t = 5 min = 5×60 = 3000 s ∆θ 2000r ev (a) ωav = = = 6.67 rev/s ∆t 3000s 1rev=2π rad ωav = 41.9 rad/s (b) Convert 2000 rev into rad ∆θ = 12560 rad Then ∆s = r∆θ ∆s = 0.3 m × 12560 rad ∆s = 3768 m Relationship between angular motion and translational motion quantities In this section, we derive some useful relationships between the angular quantities θ, ω and α of a rotating rigid object and the corresponding linear quantities s, v, and a of a point,p in the object. To do so, we must keep in mind that when a rigid object rotates around a fixed axis as in Figure 2.11, every particle of the object moves in a circle whose center is on the axis of rotation. As the particle moves along the circle through an angular displacement of θ, it moves through an arc length s. The arc length s is related to the angle θ through the equation: s =rθ (2.14) Note that in this equation the angular displacement must be expressed in rad (not degrees or revolutions). 2.2 Rotational Motion 47 Because point p, in the figure moves in a circle, the translational velocity vector → − v is always tangent to the circular path, and hence is called tangential velocity. The magnitude of the tangential velocity of the point P is by definition the tangential speed v= ∆s/∆t, where s is the distance traveled by this point measured along the circular path. Recalling that ∆s= r∆θ and noting that r is constant, we get: ∆s ∆θ v= =r (2.15) ∆t ∆t Figure 2.11 As a rigid object ∆θ rotates around the fixed axis (the Because ω = , it follows that: z axis) through O, the point P has ∆t a tangential velocity → − v that is v = ωr (2.16) always a tangent to the circular path of radius. Therefore, the tangential speed of a point on a rotating rigid object equals the radius multiplied by the angular speed. Although every point on the rigid object has the same angular speed, not every point has the same tangential speed because r is not the same for all points on the object. The tangential speed of a point on the rotating object increases as it moves outward from the center of rotation. We can relate the angular acceleration of the rotating rigid object to the tangential acceleration of the point P by taking the rate of change v. ∆v ∆ω a= =r (2.17) ∆t ∆t ∆ω Because α = , it follows that ∆t a =rα (2.18) That is, the tangential component of the translational acceleration of a point on a rotating rigid object equals the radius multiplied by the angular acceleration. Example 2.11 A rope is wrapped many times around a pulley of radius 50 cm as shown in Figure 2.12. How many revolutions of the pulley are required to raise a bucket to a height of 20 m? Solution: Figure 2.12 A rope wrapped a ∆s around a pulley of radius 50 cm. ∆θ = r 48 Unit 2 Two-dimensional motion When the bucket is raised to 20 m the same length of rope is wrapped around the pulley. Thus ∆s=20 m 20 ∆θ = 0.5 ∆θ=40 rad 1 rev=2π rad ∆θ=6.34 rev Example 2.12 The angular velocity of a bicycle wheel is 18 rad/s. If the radius of the wheel is 40 cm, what is the speed of the bicycle in m/s? Solution: r = 40 cm=0.4 m ω=18 rad/s The linear speed of the bicycle is v = ωr ∆s = 0.4 m x 18 rad/s v=7.2 m/s Example 2.13 Consider two particles, A and B, on a flat rotating disk as shown in Figure 2.13. Particle A is 20cm and particle B is 40cm from the center. The disc starts from rest and its angular speed increases to 20rad/s in 4s. (a) What is the average angular and linear acceleration for particle B? (b) What is the average angular and linear acceleration for particle A? Solution: Figure 2.13 Particle A and B on ω f − ωo the rotating disc are at different (a) αav = ∆t radius. (20 − 0)r ad /s αav = 4s 2.2 Rotational Motion 49 αav = 5r ad /s 2 a=αr a= (5 rad/s2 )(0.4 m) a= 2 m/s2 (b) The angular acceleration is the same for all particles about the axis of rotation but the linear accleration depends on r. a=αr a= (5 rad/s2 )(0.2 m) a= 1 m/s2 Discussion question 2.7 1. What is the angular speed of the second hand of a clock? What is the direction of → − ω as you view a clock hanging vertically? 2. A wheel rotates counterclockwise in the xy plane. What is the direction of → − ω ? What is the direction of → − α if the angular velocity is decreasing in time? 3. When a wheel of radius R rotates about a fixed axis, do all points on the wheel have (a) the same angular speed? and (b) the same linear speed? Exercise 2.2 1. When a wheel of radius R rotates about a fixed axis, all points on the wheel have the same angular speed. True or False. 2. Which of the following can not be a unit for angular displacement ? A. deg B. rad. rev D. rpm 3. A rope is wrapped many times around a pulley of radius 20 cm. What is the average angular velocity of the pulley if it lifts a bucket to 10 m in 5 s? 4. A particle moves in a circle 1.50 m in radius. Through what angle in radians does it rotate if it moves through an arc length of 2.50 m? What is this angle in degrees? 5. A wheel is under a constant angular deceleration of 5r ad /s 2. Its initial speed is 3 rad/s. What angular distance will it travel just before coming 50 Unit 2 Two-dimensional motion to rest? 6. A wheel initially turning at 200 rpm uniformly increases its speed to 600 rpm in 8s. Calculate: (a) the angular acceleration of the wheel in r ad /s 2. (b) the number of revolutions turned by the wheel during the 8 s interval. 2.3 Rotational Dynamics At the end of this section, you will be able to: Define the physical concept of torque in terms of force and distance from axis of rotation. Define the physical concept of moment of inertia in terms of point mass and distance from the axis of rotation. Express torque in terms of moment of inertia and angular acceleration. Solve problems involving torque and rotational kinematics. Having developed the kinematics of rotational motion, we now turn to the dynamics of rotational motion. Just as force played a big role in linear dynamics, we have a torque in rotational dynamics. We begin by defining this quantity and showing how it acts on objects in rotational motion. Next, we relate torque to our study of kinematics through an equation very similar to Newton’s second law. Torque Torque is the rotational effect of force. Torque is what causes an object to acquire angular acceleration. If F is the force acting on an object and r is the distance from the axis of rotation to the point of application of the force, as shown in Figure 2.14, the magnitude of the torque is given by: Figure 2.14 Counterclockwise rotation by F around the pivot τ = r F si nθ (2.19) point. where θ is the angle between r and F when they are drawn from the same origin. Torque is a vector quantity, meaning it has both a direction and a magnitude. Its SI unit is Nm. The direction of the torque is along the axis of rotation. It is 2.3 Rotational Dynamics 51 determined by a right-hand-rule: when you curl the fingers of your right hand in the direction of the rotation, your thumb points in the direction of the torque. Example 2.14 The object in Figure 2.15 is pivoted at O. Three forces act on it in the directions shown: F 1 = 10 N at 3.0 m from O; F 2 = 16 N at 4.0 m from O; and F 3 = 19 N at 8.0 m from O. What is the net torque about O? Solution: F 2 andF 3 give a torque in the counterclockwise direction (positive, usually) and F 1 gives a torque in the clockwise direction (negative torque). τ1 = r 1 F 1 si nθ τ1 = 3mx10N xsi n(120)0 τ1 = −25.9N m τ2 = r 2 F 2 si nθ τ2 = 4mx16N xsi n(150)0 τ2 = 32N m Figure 2.15 Three forces acting on an object pivoted at O. τ3 = r 3 F 3 si nθ τ3 = 8mx19N xsi n(45)0 τ2 = 107.4N m τnet = τ1 + τ2 + τ3 τnet = −25.9N m + 32N m + 107.4N m τnet = 113.5N m (counterclockwise direction) Moment of inertia (I) The moment of inertia of an object is the quantitative measure of rotational inertia, just as mass is the quantitative measure of linear inertia inertia in translational motion. The greater the moment of inertia of a rigid object or Figure 2.16 Point mass rotating system of particles, the greater is its resistance to change in angular velocity about O about a fixed axis of rotation. The moment of inertia depends on the mass and axis of rotation of the body.The moment of inertia is given the symbol I. For a single point mass, as shown in 52 Unit 2 Two-dimensional motion Figure 2.16, rotating at radius r from the axis of rotation the moment of inertia is: I = mr 2 (2.20) From the formula, the SI unit of moment of inertia is kg m 2. Moment of inertia is a scalar quantity. The moment of inertia for more than one particle around a fixed axis is: I = m 1 r 12 + m 2 r 22 + m 3 r 32... Example 2.15 Three particles are connected by rigid rods of negligible mass lying along the y-axis as shown in Figure 2.17. If the system rotates about the x-axis with angular speed of 2 rad/s, find the moment of inertia about the x-axis. Solution: I = m 1 r 12 + m 2 r 22 + m 3 r 32 I = 4kg (3m)2 + 2kg (2m)2 + 3kg (4m)2 I = 164 kg m 2 Figure 2.17 Three particles rotating around the x-axis. Torque and angular acceleration When a number of individual forces act on a rotating object, we can calculate the net torque: τnet = τ1 + τ2 + τ3... We can relate the net torque to angular acceleration α, by analogy with Newton’s second law of motion (F = ma). We replace m by I and a by α. τ = Iα (2.21) The angular acceleration of a rotating object is proportional to the net torque on the object. Example 2.16 2.3 Rotational Dynamics 53 When a torque of 36 Nm is applied to a wheel, the wheel acquires an angular acceleration of 24r ad /s 2. Find the rotational inertia of the wheel. Solution: τ = Iα 36N m I= 24r ad /s 2 I=1.5 kg m 2 Example 2.17 A motor capable of producing a constant torque 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm2. (a) What angular acceleration will the flywheel experience as the motor is switched on? (b) How long will the flywheel take to reach the maximum speed if starting from rest? Solution: (a) The angular acceleration is: τ α= I 100N m α= 0.1kg m 2 α = 1000 r ad /s 2 (b) The time to reach the maximum speed is: ω f = ωo + α∆t ω f − ωo 150r ad /s − 0 t= = α 1000r ad /s 2 t=0.15 s 54 Unit 2 Two-dimensional motion Exercise 2.3 1. A force of 400 N is applied to a beam at a distance of 5 m from the pivot point, as shown in Figure 2.18. Calculate the magnitude of the torque which turns the bar around pivot. 2. Three point masses, each of mass m, are placed at the corners of an equilateral triangle of side L. Find the moment of inertia of the system about an axis passing through one of the corners perpendicular to the plane of the triangle. Figure 2.18 Torque on a beam by 3. A disc with moment of inertia 2 kgm2 changes its angular speed from 400 N force. 3 rad/s to 8rad/s by a net torque of 50 Nm.How long will the disc take to change its angular speed? 2.4 Planetary motion and Kepler’s laws At the end of this section, you will be able to: Describe the motion of the planets around the Sun. State Kepler’s three laws. Apply equations to solve problems related to orbital motion. The planets orbit the Sun. They maintain their respective distances from the Sun. They do not cross each other as they revolve around the Sun. Kepler’s laws describe how planetary bodies orbit around the Sun. Kepler’s laws Figure 2.19 Earth with its Moon revolving around the Sun. Humans have observed the movements of the planets, stars, and other celestial objects for thousands of years. In early history, these observations led scientists to regard Earth as the center of the Universe. This geocentric model was elaborated and formalized by the Greek astronomer Claudius Ptolemy (c.100–c.170) in the second century and was accepted for the next 1400 years. In 1543, Polish astronomer Nicolaus Copernicus (1473–1543) suggested that Earth and the other planets revolved in circular orbits around the Sun (the 2.4 Planetary motion and Kepler’s laws 55 heliocentric model). Discussion question 2.8 Danish astronomer Tycho Brahe (1546–1601) wanted to determine how the In small group universe was constructed and pursued a project to determine the positions of discuss the both stars and planets. His observations of the planets and stars visible from following questions. Earth were carried out using only a large sextant and a compass. (The telescope What is the shape of had not yet been invented.) an orbit? What’s in the middle of the orbit? German astronomer Johannes Kepler was Brahe’s assistant for a short while What is the before Brahe’s death, where upon he acquired his mentor’s astronomical data. difference between Kepler spent 16 years trying to deduce a mathematical model for the motion of circle and ellipse? the planets. Such data are difficult to sort out because the moving planets are observed from a moving Earth. After many lengthy calculations, Kepler found that Brahe’s data on the revolution of Mars around the Sun led to a successful model. Kepler’s complete analysis of planetary motion is summarized in three statements known as Kepler’s laws. Kepler’s first law Discussion Kepler’s first law is sometimes referred to as the law of ellipses. It states that the question 2.9 orbit of a planet around the Sun is an ellipse (near circular, oval) with the Sun at Think of the planets one focus (Figure 2.20a). orbiting the Sun. Do all the planets The planet follows the ellipse in its orbit, meaning that the planet-to-Sun distance move at the same is constantly changing as the planet goes around its orbit. An ellipse is a closed speed? At which curve such that the sum of the distances from a point on the curve (r1 + r2 ) to the position are the two foci, f 1 and f 2 is constant, as shown in Figure 2.20b. planets’ orbital speeds greatest? Explain why. Figure 2.20 (a) The motion of a planet about the Sun. (b) Any distance drawn from f 1 and f 2 to a point on the curve add up to a constant. 56 Unit 2 Two-dimensional motion Kepler’s second law Kepler’s second law is sometimes referred to as the law of equal areas. It describes the speed at which any given planet will move while orbiting the Sun. Basically, it states that planets do not move with constant speed along their orbits. Instead, their speed varies so that the line joining the centers of the Sun and the planet sweeps out equal area in equal times. The point at which a planet is nearest the Sun is called perihelion. The point of greatest separation is aphelion. Figure 2.21 The shaded regions Hence Kepler’s second Law, a planet is moving fastest when it is at perihelion and shown have equal areas and slowest at aphelion. represent the same time interval. Kepler’s second law states that each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times interval. Consider Figure 2.21. The time it takes a planet to move from position A to B, sweeping out area A1 ,is exactly the time taken to move from position C to D, Discussion sweeping area A2 and to move from E to F, sweeping out area A3. These areas are question 2.10 the same: A1 =A2 =A3 In small groups discuss the Comparing the areas in the Figure 2.21 and the distance traveled along the ellipse following questions. in each case, we can see that in order for the areas to be equal, the planet must What is the period speed up as it gets closer to the Sun and slow down as it moves away. of Earth? What is meant by Kepler’s Third law the orbital period of a planet? Kepler’s third law compares the orbital period and the average radius of orbit of a Which planet has planet to those of other planets. Unlike Kepler’s first and second laws that the shortest orbital describe the motion characteristics of a single planet, the third law makes a period: Earth or Pluto? comparison between the motion characteristics of different planets. The period Is there a systematic (T ) of a planet is the time for one complete revolution around the Sun. relationship between period Kepler’s third law implies that the period for a planet to orbit the Sun increases and radius for the rapidly with the radius of its orbit. Thus we find that Mercury, the innermost planets? planet, takes only 88 days to orbit the Sun. Earth takes 365 days, while Saturn requires 10,759 days to do the same. 2.4 Planetary motion and Kepler’s laws 57 T2 Kepler’s third law states that the ratio , where T is the time period and R is the R3 average distance from the sun is the same for all planets: T2 =K (2.22) R3 K is a proportionality constant which is nearly the same for all planets. Kepler’s third law equation is valid for both circular and elliptical orbits. Notice that the constant of proportionality is independent of the mass of the planet. Therefore, the equation is valid for any planet. As an illustration, consider the orbital period and average distance from Sun (orbital radius) for Earth and Mars as given in table 2.2. Table 2.2 The orbital period and average distance from the Sun for Earth and Mars. Period(s) Average T2 /R3 (2 /m3 ) distance (m) 7 Earth 3.156 x 10 1.4957 x 1011 2.977 x 10−19 Mars 5.93 x 107 2.278 x 1011 2.975 x 10−19 Observe that the T 2 /R3 ratio is the same for Earth as it is for Mars. In fact, the T2 /R3 ratio is the same for the other planets. Example 2.18 Earth has an orbital period of 365 days and its mean distance from the Sun is 1.495×108 km. The planet Pluto’s mean distance from the Sun is 5.896×109 km. Using Kepler’s third law, calculate Pluto’s orbital period in Earth days? Solution: The given quantities are: TE = 365 d a y a, r E = 1.495 × 108 km , r P = 5.896 × 109 km We use Kepler’s third law to calculate Pluto’s orbital period. T2E T2p = R3E R3p 365days T2p 3 = 3 (1.495x108 km) (5.896x109 km) To solve for Tp , we cross-multiply and take the square root. 58 Unit 2 Two-dimensional motion Thus : TP = 9.0 × 1014 days Example 2.19 If Saturn is on average 9 times farther from the Sun than Earth is, what is this distance in Earth years? Solution: rS = 9rE , TE = 1year TS =? T2E T2S = R3E R3S 1Year T2S = R3E (9R E )3 TS = 27 years Exercise 2.4 1. According to Kepler’s laws of planetary motion, a satellite increases its speed as it approaches the Sun and decreases its speed as it moves away from the Sun. True or False. 2. Given that the Moon orbits Earth every 27.3 days and that it is an average distance of 3.84×108 m from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth’s surface. (Radius of Earth is 6380 km.) 3. How would the period of an object in a circular orbit change if the radius of the orbit doubled? A. The period would increase by a factor of 2. B. The period would decrease by a factor of 4. p C. The period would increase by a factor of 2 2. p D. The period would decrease by a factor of 2 2. https://phet.colorado.edu/sims/html/gravity-and-orbits/latest/ gravity-and-orbits_en.html 2.5 Newton’s law of universal Gravitation 59 2.5 Newton’s law of universal Gravitation At the end of this section, you will be able to: Explain what determines the strength of gravity. Brainstorming Describe how Newton’s law of universal gravitation extends our question 2.2 understanding of Kepler’s laws. Imagin the Sun’s Apply equations to solve problems related to Newton’s law of universal gravity is suddenly switched off. What gravitation. will happen to the planets? Planets orbit the Sun. If we look more closely at the Solar System, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity. All these motions are governed by gravitational force. Galileo Galilei (1564-1642) pointed out that heavy and light objects fall toward Earth at the same rate (so long as air resistance is the same for each). But it took Sir Isaac Newton (in 1666) to realize that this force of attraction between masses is universal. Discussion Question 2.11 Newton proved that the force that causes, for example, an apple to fall toward the ground is the same force that causes the Moon to fall around, or orbit, Earth. This What keeps the planets in orbit? universal force also acts between the Earth and the Sun, or any other star and its Explain your satellites. Each attracts the other. answer. Newton defined this attraction mathematically. The force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Gm 1 m 2 Fg = (2.23) r2 where G is a constant, called the universal gravitational constant, m1 is the first mass, m2 is the second mass, and r is the distance between the two masses. G is a Figure 2.22 Gravitational universal constant, meaning that it is thought to be the same everywhere in the attraction is along a line joining Universe. The value G in SI units is G = 6.67x10−11 Nm2 /kg2. the centers of mass of the two bodies. 60 Unit 2 Two-dimensional motion The gravitational force is always attractive, and it depends only on the masses involved and the distance between them. The force is directed along the line joining the two masses, as shown in Figure 2.22. The magnitude of the force on each body is the same but the direction is opposite, consistent with Newton’s third law, action-reaction (F12 = -F21.) Example 2.20 A 10 kg mass and a 100 kg mass are 1 meter apart. What is the force of attraction between them? Solution: G is given above, m1 = 10 kg, and m2 = 100 kg. Putting these values into Newton’s gravitational force Gm 1 m 2 Fg = r2 ³ ´ 6.673x10−11 N m 2 2 x10kg x100kg kg Fg = (1m)2 Fg = 6.67 x 10−8 N Example 2.21 If a person has a mass of 60.0 kg, what would be the force of gravitational attraction on him at Earth’s surface? Solution: G is given above, Earth’s mass ME is 5.97 × 1024 kg, and the radius rE of Earth is 6.38 × 106 m. Putting these values into Newton’s gravitational law: G ME m Fg = rE 2 (6.673x10−11 N m 2 /kg 2 )x5.97x1024 kg x60kg Fg = 2 (6.38x106 m) F g = 584N We can check this result with the relationship: Fg = mg Fg = 60 kgx9.8 m/s2 Fg = 588 N 2.5 Newton’s law of universal Gravitation 61 You may remember that g, the acceleration due to gravity, is another important Discussion constant related to gravity. By substituting g for a in the equation for Newton’s question 2.12 second law of motion (F = ma) we get: Fg = mg. By what factor Combining this with the equation for universal gravitation force gives would a person’s weight at the surface G ME m mg = (2.24) of Earth change rE 2 if Earth had its Cancelling the mass m on both sides of the equation and filling in the values for present mass but the gravitational constant, mass and radius of the Earth, gives the value of g on eight times its the surface of the earth. which may look familiar. present volume? By what factor −11 2 2 24 (6.673x10 N m /kg )x5.97x10 kg would a person’s g= 2 = 9.8 m/s 2 (2.25) 6 weight at the surface (6.38x10 m) of Earth change Centripetal Force if Earth had its present size but only It is possible to derive Kepler’s third law from Newton’s law of universal gravitation. one-third its present A force that pulls an object towards the centre of a circle is called centripetal force mass? as shown in Figure 2.23. The source for the centripetal force in the Solar System is the gravitational force of the Sun. Without the centripetal force from the Sun the planets would travel in a straight line. The velocity of the planets is high enough so that they continuously accelerate towards the Sun without ever leaving their orbits. It is for this reason that the planets do not fall into the sun from its strong gravitational force of attraction. Applying Newton’s second law of motion to circular motion gives an expression for centripetal force. mv 2 Fc = (2.26) r where v is the tangential speed and r is the radius of the orbit and m is mass of the planet. The gravitational attraction of the Sun provides the centripetal force needed to keep planets in orbit around the Sun. Earth’s gravity keeps the Moon and all types Figure 2.23 Centripetal force of satellite in orbit around Earth. Because the gravitational force provides the constantly pulls the object towards centripetal acceleration of the planet, it follows that: the center of the circle. mp v 2 G Ms mp = (2.27) r r2 Mp is mass of the planet, Ms is mass of the sun (Ms ≈1.989x1030 kg) and v is the 62 Unit 2 Two-dimensional motion speed of the plane about the sun. G Ms v2 = (2.28) r The orbital speed of the planet is 2πr v= (2.29) T where T is the period of the planet about the Sun. Thus (2πr )2 G M s = (2.30) T2 r T2 4π2 = = 2.97x10−19 r 3 G Ms This equation is Kepler’s third law: the square of the period is proportional to the cube of the distance of the planet from the Sun. The proportionality constant K takes the value: 4π2 K= ≈ 2.97x10−19 G Ms The above equation is therefore valid for any planet. If we were to consider the orbit of a satellite such as the Moon about the Earth, the constant would have a 4π2 different value, with the Sun’s mass replaced by the Earth’s mass; that is,. G ME Exercise 2.5 1. The gravitational force between a 60 kg man and Earth is not equal because Earth is more massive than the man therefore, it exerts the greatest force. True or False. 2. Two objects are attracted to each other by a gravitational force F. If the distance between the objects is doubled, what is the new gravitational force between the objects in terms of F ? A. 4 F B. 1/4F C. 16F D. 1/16F 3. Newton’s law of gravitation applies to: A. Small bodies only. B. Plants only. C. All bodies irrespective of their size. D. Moon and satellites only 4. Suppose the gravitational force between two spheres is 30 N. If the 2.5 Newton’s law of universal Gravitation 63 magnitude of each mass doubles, what is the force between the masses? 5. Calculate the mass of the Sun, noting that the period of Earth’s orbit around the Sun is 3.156 x 107 s and its distance from the Sun is 1.496 x 1011 m. 6. A hypothetical planet has a mass of four times that of the Earth and radius of twice that of the Earth? What is the acceleration due to gravity on the planet in terms of the acceleration on Earth? https://phet.colorado.edu/sims/html/gravity-force-lab/latest/ gravity-force-lab_en.html Unit summary Projectile refers to an object that is in flight with acceleration due to gravity after being thrown or projected. A football kicked in a game, a bullet fired from a gun, the flight of a golf ball, a jet of water escaping a hose are a few common examples of projectile motion The horizontal component of the velocity is constant throughout the projectile motion. The vertical motion has a constant acceleration which is the accleration due to gravity. In projectile motion the time to cover both the horizontal and vertical displacement is the same. When the angle of projection is measured with the horizontal axis: given by: – The vertical displacement is: 1 ∆ y = v 0 sin θ t + 2 g t2 – The horizontal displacement is: ∆ x=v 0 cosθt 64 Unit 2 Two-dimensional motion – The vertical velocity is: v y =v 0 sinθ+gt – The horizontal velocity is: v x =v 0 cosθ When a rigid object rotates about a fixed axis, the angular position, angular speed, and angular acceleration are related to the translational position, translational speed, and translational acceleration through the relationships s = rθ , v = rω, a= rα For a body rotating around a fixed axis, every particle on the body has the same rotational quantities ∆θ, ω, and α. That is ∆θ, ω, and α describe the rotational motion of the entire body. ω and α are vector quantities. The direction of → − ω is given by the right-hand-rule (RHR) and the → − direction of α follows from its definition: α = ∆ω ∆t Right-hand rule: Wrap your four right-hand fingers in the direction of rotation. Your extended thumb points in the direction of → − ω. Mathematically, we have defined the rotational quantities θ, ω, and α similar to how we defined the linear quantities s, v, and a for linear motion. Therefore, the rotational equations of motion with constant angular acceleration, should also be similar. The speed at which any planet moves through space is constantly changing. A planet moves fastest when it is closest to the Sun and slowest when it is furthest from the Sun. Kepler’s laws apply to any celestial body orbiting any other celestial body. For example, any planet around a Sun, the Moon around Earth, any satellite around Earth. A planet in the Solar System is in orbit around the Sun, due to the gravitational force on the planet exerted by the gravitational force of the Sun. 2.5 Newton’s law of universal Gravitation 65 Every object in the Universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Torque is the rotational effect of force. Moment of inertia is a measure of an object’s resistance to changes to its rotation. The motion of planets around the Sun governed by the gravitational force the Sun and the planets.