There seems to be three different groups of students: A group around 6 A group around 12 A group around 16

10 5 0 0 5 10 15 20 25 30 There seems to be three different groups of students: A group around 6 A group around 12 A group around 16 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 55 / 67

Trajectory of a Football z x, y v 0 Assume that you hit a football lying on the ground. It s initial speed is v 0 making an angle with the ground. Choose the origin of time such that t i = 0 and origin of coordinate axis such that x(0) = 0 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 56 / 67

Trajectory of a Football Third Week x, y z v 0 x(t) = v 0 t + 1 2 gt2 ẑ z(t) = v 0z t + 1 2 gt2 z(t) = 0 when t = 0(initial time) and at t = 2v 0z g (when the ball hits the ground) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 56 / 67

Trajectory of a Football Third Week z x, y v 0 The flight time of the ball is t = 2v 0z g. The only acceleration is along the z axis. v 0z g is the time it take for the z component of the velocity to become zero, i.e. the time it takes to reach maximum height The time it takes to fall down is the same (in the absence of air friction) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 56 / 67

Trajectory of a Football Third Week z x, y v 0 Assume that x and y axis are chosen such that v = v 0 sin ẑ + v 0 cos ˆx v y (t) = 0, y(t) = 0 for all times x(t) = v 0x t. Range is the distance the ball covers during its flight, i.e. R = x(t f ) ( R = v 0x 2v ) 0z g ( = v 0 cos 2( v ) 0 sin ) g = v 2 0 sin 2 g (29) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 56 / 67

Trajectory of a Football z x, y v 0 R = v 2 0 sin(2) g sin 2 has maximum value of 1 when = 45 Increasing v 0 by a factor of 2 increases the range by 4. If 1 + 2 = π 2, their ranges are the same Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 56 / 67

Trajectory Third Week Trajectory is a relationship between the components of the position of a particle that does not involve time. When the particle is at the horizontal distance x, the time that has passed is t(x) = x/v 0x. The z coordinate of the particle at that time is z(x) = v 0z t(x) + 1 2 gt(x)2 ( ) x = v 0z + 1 ( ) x 2 v 0x 2 g v 0x g = 2v0 2 x(x R) (29) cos2 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 57 / 67

Trajectory Third Week Trajectory is a relationship between the components of the position of a particle that does not involve time. When the particle is at the horizontal distance x, the time that has passed is t(x) = x/v 0x. The z coordinate of the particle at that time is z(x) = v 0z t(x) + 1 2 gt(x)2 ( ) x = v 0z + 1 ( ) x 2 v 0x 2 g v 0x g = 2v0 2 x(x R) (29) cos2 Trajectory v v 0 0y z v 0x v 0x v 0x x v 0y v 0 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 57 / 67

Example Third Week v 0 P(x 0, y 0 ) Q: What is the distance P? θ Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example Q: What is the distance P? v 0 θ P(x 0, y 0 ) To find the point P, we will use the fact that point P is both on the parabola describing the trajectory, and also on the line that describes the hill. Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example y x v 0 P(x 0, y 0 ) Q: What is the distance P? First choose a coordinate axis. θ Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example y θ v 0 a x a y x P(x 0, y 0 ) Q: What is the distance P? First choose a coordinate axis. The initial velocity and acceleration in these new coordinate axes are: v 0 = v 0 cos(θ )ˆx + v 0 sin(θ )ŷ v 0y θ v 0x a = g cos ( ŷ) + g sin ( ˆx) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example Third Week v 0y y θ v 0 a x θ v 0x a y x P(x 0, y 0 ) Q: What is the distance P? The velocity at time t can be obtained as v(t) = v 0 + t 0 a(t )dt = v 0 + t a = [v 0 cos(θ ) gt sin ] ˆx + [v 0 sin(θ ) gt cos ] ŷ Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example Third Week v 0y y θ v 0 a x θ v 0x a y x P(x 0, y 0 ) Q: What is the distance P? The velocity at time t can be obtained as v(t) = v 0 + t 0 a(t )dt = v 0 + t a = [v 0 cos(θ ) gt sin ] ˆx + [v 0 sin(θ ) gt cos ] ŷ The position at time t is t r(t) = r 0 + v(t )dt 0 (30) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example Third Week v 0y y θ v 0 a x θ v 0x a y x P(x 0, y 0 ) Q: What is the distance P? The position at time t is t r(t) = r 0 + v(t )dt 0 [ = v 0 t cos(θ ) 1 ] 2 gt2 sin ˆx [ + v 0 t sin(θ ) 1 ] 2 gt2 cos ŷ Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example y θ v 0 a x a y x P(x 0, y 0 ) Q: What is the distance P? Hence, if the object reaches the point P at time t 0, x 0 = v 0 t 0 cos(θ ) 1 2 gt2 0 sin v 0y θ v 0x y 0 = v 0 t 0 sin(θ ) 1 2 gt2 0 cos (30) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example v 0y y θ v 0 a x θ v 0x a y x P(x 0, y 0 ) Q: What is the distance P? At point P, y 0 = 0 v 0 t 0 sin(θ ) 1 2 gt2 0 cos = 0 (30) which has solutions t 0 = 0 or t 0 = 2v 0 sin(θ ) g cos Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example v 0y y θ v 0 a x θ v 0x a y x P(x 0, y 0 ) Q: What is the distance P? At point P, y 0 = 0 v 0 t 0 sin(θ ) 1 2 gt2 0 cos = 0 (30) which has solutions t 0 = 0 or t 0 = 2v 0 sin(θ ) g cos t 0 = 0 is the beginning of motion. The second solution is the solution we are looking for. Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Example v 0y y θ v 0 a x θ v 0x a y x P(x 0, y 0 ) Q: What is the distance P? The distance P = x 0. Using t 0 = 2v 0 sin(θ ) g cos x 0 = v 0 ( 2v0 sin(θ ) g cos 1 2 g ( 2v0 sin(θ ) g cos ) sin(θ ) ) 2 cos (30) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 58 / 67

Relative Motion Third Week y From the definition of vector addition r = R + r. y r x P The displacement of the point P in a time t is R r r = R + r (31) The velocities in the two reference frames are related by v = v + V x Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 59 / 67

Relative Motion Third Week y y P r x R r x From the definition of vector addition r = R + r. The displacement of the point P in a time t is r t = R + r t t The velocities in the two reference frames are related by v = v + V (31) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 59 / 67

Relative Motion Third Week y y From the definition of vector addition ASSUMPTINS r = R + r. r and r are measured P The in different displacement of the point P in a reference r frames. We aretime assuming t is that they are equal. Furthermore, we are assuming the t is the same in r x t = R + r (31) both reference frames. t t R r The velocities in the two reference frames are related by v = v + V x Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 59 / 67

Example Question 3.78 Raindrops make an angle θ with the vertical when viewed through a moving train window. If the speed of the train is v T, what is the speed of the raindrops in the reference frame of the Earth in which they are assumed to fall vertically? Solution: Let v R = vẑ be the speed of the raindrops in the reference frame of Earth, v E be the velocity of the Earth relative to the train, i.e. v E = v T. z v E v R x Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 60 / 67

Example Third Week Question 3.78 Raindrops make an angle θ with the vertical when viewed through a moving train window. If the speed of the train is v T, what is the speed of the raindrops in the reference frame of the Earth in which they are assumed to fall vertically? Solution: Let v R = vẑ be the speed of the raindrops in the reference frame of Earth, v E be the velocity of the Earth relative to the train, i.e. v E = v T. z θ v RT v E v R x Let v RT be the velocity of the raindrops relative to train Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 60 / 67

Example Question 3.78 Raindrops make an angle θ with the vertical when viewed through a moving train window. If the speed of the train is v T, what is the speed of the raindrops in the reference frame of the Earth in which they are assumed to fall vertically? Solution: Let v R = vẑ be the speed of the raindrops in the reference frame of Earth, v E be the velocity of the Earth relative to the train, i.e. v E = v T. z v RT v E θ v R x It is given that v RT makes θ radians with respect to the vertical From the figure, it is seen that tan θ = v E v R = v R = v T cot θ (32) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 60 / 67

Reference Frames event: position+time A reference frame is a coordinate axis (to measure the position of an event) And a clock at each point of space (to measure the time of an event) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 61 / 67

Dynamics-Newton s Laws of Motion 1 st Law: In an inertial reference frame, in the absence of any external influences, the velocity of an object is constant Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 62 / 67

Dynamics-Newton s Laws of Motion 1 st Law: In an inertial reference frame, in the absence of any external influences, the velocity of an object is constant This is a definition of an inertial reference frame Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 62 / 67

Dynamics-Newton s Laws of Motion Inertial Reference Frame To test if a given reference frame is inertial, consider a test object Eliminate all external influecens. Check to see if the object accelerates or not If the object is not accelerating, that reference frame is an inertial reference frame Given one inertial reference frame, any other frame that moves at constant velocity relative to the inertial reference frame is inertial: v = V + v (33) If a given reference frame is an inertial reference frame, all objects obey Newtons 1 st law in that frame Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 63 / 67

Dynamics-Newton s Laws of Motion 2 nd Law: In an inertial reference frame, the acceleration of an object is proportional to the force acting on the object. The proportionality constant is 1 m where m is the mass of the object a = F m (34) 3 rd Law: If an object A exerts a force F AB on another object B, then object B also exerts a force F BA on object A whose magnitude is equal to the magnitude of F AB, but opposite in direction: F AB = F BA (35) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 64 / 67

Dynamics-Newton s Laws of Motion 2 nd and 3 rd laws define the mass of an object By the 3 rd law, the magnitudes of the force acting on the standard mass and the unknown mass are equal: Using 2 nd law: ma = m s a s (36) Accelerations can be measured experimentally. Hence the unknown mass can be obtained as: m a m s a s m = m s a s a (37) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 65 / 67

Dynamics-Newton s Laws of Motion nce the mass is defined, 2 nd Law can be considered as the definition of the force. Also, if the force is given (by some means), the second law can be used to obtain acceleration. Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS109 66 / 67