Dynamics-Newton s Laws of Motion

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1 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 PHYS / 78

2 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 PHYS / 78

3 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 = a = A + a (33) If a given reference frame is an inertial reference frame, all objects obey Newton s 1 st law in that frame Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

4 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 PHYS / 78

5 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 PHYS / 78

6 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 PHYS / 78

7 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 PHYS / 78

8 Once 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. Unit of Force: [ F] = [m a] = [m][ a] = kg m N(Newton) (38) s2 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

9 Once the mass is defined, 2 nd Law can be considered as the definition of the force. Also, if the ATTENTION force is given (by some means), the second law can be used to obtain There is acceleration. no force due to acceleration! Unit of Force: The force is the cause of acceleration! [ F] = [m a] = [m][ a] = kg m N(Newton) (38) s2 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

10 Weight m w Weight, w, is the force acting on an object due to gravity. Near Earth, all object accelerate with the same acceleration g. By Newton s second law, the force acting on an object of mass m is w = m g (39) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

11 Example 1: Mass on a scale N m N: unknown force acting on the body by the scale w: force of gravity acting on the body scale w Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

12 Example 1: Mass on a scale N m N: unknown force acting on the body by the scale w: force of gravity acting on the body Free body diagram: A diagram of masses only with the forces acting on each body shown separately w Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

13 Example 1: Free Body Diagram N m Object is not accelerating: F net = N + w 0 = N = w (40) The scale shows the magnitude of the force acting on it: N = w = mg w Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

14 Example 2: Mass on a scale inside an Elevator z a e is the acceleration of the elevator The vectors in the problem are: N m scale w a e N = Nẑ(Unknown) (41) w = mgẑ (42) a e = a e ẑ (43) (In drawing the figure, it is assumed that a e < 0) If the mass m stays on the scale, a = a e = a e ẑ Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

15 Example 2: Mass on a scale inside an Elevator z N m scale a e Net force acting on the mass: F T = (N mg)ẑ = ma e ẑ = N =m(g + a e ) (41) The force acting on the scale is N, Scale will show a weight m(g + a e ). w Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

16 14 th Century Bologna University Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

17 Learners and Learning Herb Simon Nobel laureate, Social Scientist, one of the founders of AI Learning results from what the student does and thinks and only from what the student does and thinks. The teacher can advance learning only by influencing what the student does to learn. Dylan William renowned UK expert on maths education... teachers do not create learning, and yet most teachers behave as if they do. Learners create learning. Teachers create the conditions under which learning can take place. Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

18 Example 3: Mass on an Inclined Plane The object on the inclined surface N A block sits on a frictionless incline as shown in the figure The forces acting on the mass are its weight and the normal force O α w Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

19 Example 3: Mass on an Inclined Plane Free Body Diagram N w Free body diagram includes only the mass and the forces Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

20 Example 3: Mass on an Inclined Plane N O α F T N w The net force has to be along the surface of the inclined plane Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

21 Example 3: Mass on an Inclined Plane y N x In terms of their components, the forces can be written as: N = Nŷ (Unknown) (42) F T N α w w = mg cos αŷ mg sin αˆx (43) The net force is: F T = (N mg cos α)ŷ mg sin αˆx (44) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

22 Example 3: Mass on an Inclined Plane y N F T N α w x The net force is: F T = (N mg cos α)ŷ mg sin αˆx (42) The acceleration along the y direction should be zero, hence a y = 0 = F Ty = 0 N mg cos α = 0 = N = mg cos α (43) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

23 Example 3: Mass on an Inclined Plane y N x The net force is: F T N α F T = mg sin αˆx (42) Using Newton s second law: w a = F T m = g sin αˆx (43) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

24 Example 3: Mass on an Inclined Plane N a g O α w Figure : a = g sin α π 2 α Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

25 Tension of a String T F F Tension is the magnitude of the force acting on a string In the above figure, if the string has negligible mass (m = 0), than F T = F + F = m a = 0 (42) A massless string transfers force along its length without changing its magnitude. Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

26 Example: Two masses attached by a massless string T m 1 m 2 F Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

27 Example: Two masses attached by a massless string N 1 N 2 F 1 F2 F w 1 w 2 No motion in the vertical direction: w 1 + N 1 = 0 and w 2 + N 2 = 0 The string is massless ( F 1 ) + ( F 2 ) = 0 = F 2 = F 1 If the elasticity of the string is neglected, both masses should have the same acceleration: F 1 = m 1 a, F + F 2 = m 2 a Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

28 Example: Two masses attached by a massless string N 1 N 2 F 1 F2 F w 1 w 2 Since all the forces and accelerations are in the horizontal direction, I will only write the horizontal components of each vector } F 1 F F = (m 1 + m 2 )a = a = F + F 2 = m 1 a = F F 1 = m 2 a m 1 + m 2 (43) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

29 Atwood s Machine Third Week m 1 m 2 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

30 Atwood s Machine Third Week z T 1 m 1 T 2 m 2 w 1 w 2 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

31 Atwood s Machine z Let accelerations be a 1 = a 1 ẑ, and a 2 = a 2 ẑ. T 1 = T ẑ, T 1 = T ẑ where T is the (unknown) tension of the string a 1 = a 1 ẑ, a 2 = a 2 ẑ (a i s are unknown) T 1 m 1 w 1 m 2 w 2 T 2 w 1 = m 1 gẑ, w 2 = m 2 gẑ For the masses m 1 and m 2 : T i + w i = m i a i T m i g = m i a i The velocities of the masses have to have equal magnitudes but opposite direction: v 1 = v 2 a 1 = a 2 Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

32 Atwood s Machine Third Week z a 2 = a 1 (44) T m 1 g = m 1 a 1 (45) T m 2 g = m 2 a 2 (46) T 1 m 1 w 1 m 2 w 2 T 2 Subtracting the second equation from the third and using the first: (m 1 m 2 )g = m 2 a 2 m 1 a 1 = (m 2 +m 1 )a 1 = a 1 = m 1 m 2 m 1 + m 2 g (47) Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

33 Accelerometer Third Week A ball of mass m is suspended from a point by a massless string. m Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

34 Accelerometer θ v(t) A ball of mass m is suspended from a point by a massless string. If the suspension point starts to move with constant acceleration a, what is the relation between the angle θ and a a? Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

35 Accelerometer Third Week y α T x Once the oscillations settle down, the acceleration of the ball is equal to the acceleration of the suspension point aˆx w Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

36 Accelerometer Third Week y α T w = mgŷ, T = T cos αŷ + T sin αˆx F T = (T cos α mg)ŷ + T sin αˆx (48) By Newton s second law: F T = maˆx: x T cos α mg = 0 = T = mg cos α (49) w T sin α = ma = mg tan α = ma (50) Hence tan α = a g Altuğ Özpineci ( METU ) Phys109-MECHANICS PHYS / 78

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 10 5 0 0 5