Units are important anyway

Transcription

1 Ch. 1 Units -- SI System (length m, Mass Kg and Time s). Dimensions -- First check of Mathematical relation. Trigonometry -- Cosine, Sine and Tangent functions. -- Pythagorean Theorem Scalar and Vector Quantities Scalar can be described completely by magnitude Vector needs direction along with magnitude to be fully described. Vector Addition and Subtraction Vector Components Method Units are important anyway

2 Vector Components Any vector can be resolved into two components x-component Along x-axis (Horizontal) y-component Along y-axis (Vertical) A = 2 km at 30 o North of East y-axis θ A A x A y X-axis A x = A cosθ Along x-axis km A = 2 km θ = 30 o A y = A Sinθ Along y-axis 1.0 km

3 Addition of Vectors by Means of Components A = 2 km 30 o North of East B = 3 km 60 o North of East A x = 2 km Cos (30 o ) B x = 3 km Cos (60 o ) A y = 2 km Sin (30 o ) B y = 3 km Sin (60 o ) R x = A x + B x R y = A y + B y R = [R 2 x + R 2 y ] 1/2 θ = Tan 1 R ( R y x ) 4.8 km 48 o

4 Ch. 2 Units in MKS system? Distance Change of position m Displacement Distance with direction m Scalar or Vector? Scalar Vector Speed Rate of change of position m/s Scalar Velocity Speed with direction m/s Vector Acceleration Rate of change of velocity m/s 2 Vector Equations of motion: v = v 0 + a t x= v 0 t + ½ a t 2 x x0 v = = t t 0 x t 1 x = v + v ) t 2 ( 0 2 a x= v 2 v 0 2

5 Remember Applications of Eqs of motions Decide positive and negative directions Stick to the signs of directions throughout the problem Write down what you know and what you need to find Make free body diagram Starting from rest => v 0 = 0 Deceleration ---not necessarily negative velocity More than two answers --- use common sense All the equations of motion we have are good only for constant acceleration If acceleration changes Divide motion into segments --- final velocity of one segment will be the initial velocity of the next segment

6 Ch. 3 In 2-Dimension, x and y components of the motion are dealt independent of each other. Equation of motions are same.

7 Ch. 4 Newton s Laws 1 st Law 2 nd Law A net non zero force is required to change the velocity of an object. What happens when there is a net non-zero force applied? The results is acceleration F a = F = a OR m m 3 rd Law Where the force/s coming from? In 2 dimension F x = ma F = x & y ma y Equilibrium a = 0 x = 0 F y F & = 0

8 F M Em = G W = mg 2 r Fundamental Forces 1. Gravitational 2. Strong Nuclear 3. Electro-weak Frictional Force f µ MAX s Non-Fundamental Result of fundamental Forces Frictional force Normal force Tension Centripetal force f = µ F = sfn k k N

9 Ch. 5 Uniform Circular Motion v r = 2π T Centripetal Force FF cc = mmvv2 rr Centripetal acceleration 2 v a c = r Net force towards the center Friction is responsible to provide necessary centripetal force when negotiating unbanked curve. On banked curve horizontal component of normal force can provide necessary centripetal force : don t have to depend on friction Gravitational force provide required centripetal force for satellites in a circular orbits around the earth.

10 Ch. 6 KE = ½ mv 2 PE = mgh E = KE+ PE W = ΔKE Joule Conservative and non-conservative forces Principle of Energy Conservation Energy can not be created or destroyed but it can transform from one form to another form W P = = Fv t Joule/s = watts

11 Exam#1 Friday, March 01 Ch. 1 to Ch. 6 In Exam No notes or book. You can use calculator. Formula sheet will be given with the exam.