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9TH CLASSES PHYSICS DAILY PLAN SUBJECT: CIRCULAR MOTION GOALS: DURATION: IN PRACTICE: a=V/t=V.V /r a= a=(w.r)2/r a= w2 .r V2 r The acceleration vector in this case perpendicular to the path and always points toward the center of the circle. a PRESENTATION: a a UNIFORM CIRCULAR MOTION : The motion of an object that moves on a circle with constant speed is called uniform circular motion. V a An acceleration of this nature is called a centripetal acceleration or radial acceleration. Centripetal Force : V In uniform circular motion the object is under the action of a force which is always directed toward the center of the circle. This force keeps the object moving in the circle with a constant speed. Period (T) : The time to complete one turn is called the period of the motion. These forces ara called centripetal forces. By appliying Newton’s 2nd Law Frequency (f) : The number of turns in one second is called the frequency of the motion and it is equal to ; f=1/T or T.f=1 PERIOD (T) FREQUENCY (f) SECOND (s) 1/s or s-1 Fc = m.a Fc = m. O V r V V Linear Speed (V) : The distance travelled by the object during the period (T) is S=2r then; V= S/T V=2r/T V= 2fr Angular Speed (w) : The angle swept by the radius line in unit time is called the angular speed (w). If the angle swept is in time t, then ; w=/t We know that the radius sweeps 2 radians during one complete turn. Thus ; w=2/T w=2f V2 Fc = m.w2.r r V V F F F F V V TYPICAL PROBLEMS : An object moving around a horizantal circle ; V T V=2r/T w=2/T r V= w.r Centripetal Accleration : There are two ways in which an acceleration can be produced; by a change in the magnitude of the velocity & by a change in the direction of the velocity. V1 V1 A r O B A B r r V2 V1 D O r L Fc = m.a V V2 E If is small, the length of the arc AB will approximately be equal to line AB. Thus; AB = V.t V/V=(V.V)/r An object moving around a verticle circle ; V AOBDBE V/V = AB /r Ex: A 0,5kg mass is attached to a string 80cm long and it swings in a horizantal circle with a speed of 4m/s. a) Find the centripetal acceleration of the mass. b) Find the tension in the string. (a:20m/s , b: 10N ) V m Fc = m.a V2 T= m. r T T M T W W T W K N W V2 At K T+W.cos= m. r 2 V At L T+W= m. r 2 V At M T= m. r V2 At N T-W= m. r Ex: A 0,2kg stone is attached to a stribg 50cm long. It swings in a vertical circle at an angular frequency of f= 4rev/s a) Find the centripetal acc. b) Find the tension in the string when the stone is at the highest point. c) At the highest point, when it is 180cm above the ground the string breakes and the stone flies off. Find the horizantal distance that the stone travels before it strikes the ground. (=3,g=10m/s2) A car turning an unbanked curved road ; V2 r V2 .m.g m. r N Ff m. W V .gr Ff r Ex: A 600kg car travels around a circular path of radius r=100m at uniform speed in 20s. a) What must be the minimum coefficient of friction between the tires and the road for the car to remain on the circular path? b) Find the centripetal force acting on the car? (=3) A car turning a banked curved road ; 1. If there is no friction y N x r Ny N x W W V V2 Nx = m. N.sin = m. r r 2 V= r.g.tan N.cos = m.g Ny = m.g 2. N If there is a friction ; N N x Ff Ny inclined plane Nx + Ffx = m. V2 r N.sin+ Ff.cos= m. W V2 r Ex: A 1000kg car rounds a curved banked road of radius 75m at a speed 15m/s. Find the minimum angle of banking for the car to remain on the road. ( Assume that there is no friction and =3 ) HOMEWORK: MULTIMEDIA: DEMONSTRATION: EXPERIMENT: TEACHER: DIRECTOR: V