Presentation on theme: "7.1 Characteristics of Uniform Circular Motion Objectives"— Presentation transcript:
17.1 Characteristics of Uniform Circular Motion Objectives Speed and VelocityAccelerationThe Centripetal Force RequirementMathematics of Circular MotionHomework: castle learning
2Uniform circular motion Uniform circular motion is the motion of an object in a circle with a constant or uniform speed. The velocity is changing because the direction of motion is changing.Velocity:magnitude: constantDirection: changing, tangent to the circle,
3Velocity is TANGENT to the Time for one revolution Circular VelocityObjects traveling in circular motion have constant speed and constantly CHANGING velocity – changing in direction but not magnitudeHow do we define VELOCITY?Velocity is TANGENT to thecircle at all pointsWhat ‘t’ are we talking about?What ‘d’ are we talking about?PERIOD (T)Time for one revolutionCIRCUMFERENCEC = 2πr = πd
4Speed (tangential speed) and velocity Speed, magnitude of velocity:Direction of velocity:Tangent to the circle.
5vavg =2·π·RTThe average speed and the Radius of the circle are directly proportional. The further away from the center, the bigger the speed.
6Example 1A vehicle travels at a constant speed of 6.0 meters per second around a horizontal circular curve with a radius of 24 meters. The mass of the vehicle is 4.4 × 103 kilograms. An icy patch is located at P on the curve. On the icy patch of pavement, the frictional force of the vehicle is zero. Which arrow best represents the direction of the vehicle's velocity when it reaches icy patch P?abcd
7Example 2The diagram shows a 2.0-kilogram model airplane attached to a wire. The airplane is flying clockwise in a horizontal circle of radius 20. meters at 30. meters per second. If the wire breaks when the airplane is at the position shown, toward which point where will the airplane move? (A, B, C, D)
8Example 3The diagram shows an object with a mass of 1.0 kg attached to a string 0.50 meter long. The object is moving at a constant speed of 5.0 meters per second in a horizontal circular path with center at point O.If the string is cut when the object is at the position shown, draw the path the object will travel from this position.
9AccelerationAn object moving in uniform circular motion is moving in a circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction.Since the velocity is changing. The object is accelerating.where vi represents the initial velocity and vf represents the final velocity after some time of t
10Example 4The initial and final speed of a ball at two different points in time is shown below. The direction of the ball is indicated by the arrow. For each case, indicate if there is an acceleration. Explain why or why not. Indicate the direction of the acceleration.a.No, no change in velocityyes, change in velocity, rightb.yes, change in velocity, leftc.yes, change in velocity, leftd.
11Direction of the Acceleration Vector in uniform circular moiton -+The velocity change is directed towards point C - the center of the circle.The acceleration of the object is dependent upon this velocity change and is in the same direction as this velocity change. The acceleration is directed towards point C, the center as well, this acceleration is called the centripetal acceleration.
12Example 5An object is moving in a clockwise direction around a circle at constant speed.Which vector below represents the direction of the velocity vector when the object is located at point B on the circle?Which vector below represents the direction of the acceleration vector when the object is located at point C on the circle?Which vector below represents the direction of the velocity vector when the object is located at point C on the circle?Which vector below represents the direction of the acceleration vector when the object is located at point A on the circle?dbad
13The Centripetal Force Requirement According to Newton's second law of motion, an object which experiences an acceleration requires a net force.The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration. This is sometimes referred to as the centripetal force requirement.The word centripetal means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center.
14Centripetal Force Inertia causes objects to travel STRAIGHT Paths can be bent by FORCESCENTRIPETAL FORCE bends an object’s path into a circle - pulling toward the CENTER
16Inertia, Force and Acceleration for an Automobile Passenger Observe that the passenger (in blue) continues in a straight-line motion until he strikes the shoulder of the driver (in red). Once striking the driver, a force is applied to the passenger to force the passenger to the right and thus complete the turn. This force is the cause for turning. Without this force, the passenger would still travel in a straight line at constant speed (inertia).
18Without a centripetal force, an object in motion continues along a straight-line path. With a centripetal force, an object in motion will be accelerated and change its direction.
19Centrifugal force is a not real centrifugal (center fleeing) forceA ‘fictitious’ force that is experienced from INSIDE a circular motion systemcentripetal (center seeking) forceA true force that pushes or pulls an object toward the center of a circular path
20Example 6An object is moving in a clockwise direction around a circle at constant speedWhich vector below represents the direction of the force vector when the object is located at point A on the circle?Which vector below represents the direction of the force vector when the object is located at point C on the circle?Which vector below represents the direction of the velocity vector when the object is located at point B on the circle? Which vector below represents the direction of the velocity vector when the object is located at point C on the circle?Which vector below represents the direction of the acceleration vector when the object is located at point B on the circle?dbdac
21Mathematics of Uniform Circular Motion vavg =2·π·RTa =v2Ra =4π2RT2
22Relationship between quantities v2RFnet and a is directly proportional to the v2.F ~ v2a ~ v2If the speed of the object is doubled, the net force required for that object's circular motion and its acceleration are quadrupled.a is not related to massF ~ m
23Example 1 F = m v2 / r F in directly proportional to v2 A car going around a curve is acted upon by a centripetal force, F. If the speed of the car were twice as great, the centripetal force necessary to keep it moving in the same path would beF2F F/24F F = m v2 / rF in directly proportional to v2F in increased by 22
24Example 2Anna Litical is practicing a centripetal force demonstration at home. She fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle? Explain using an equation.It will require more force to accelerate a half-full bucket of water compared to a quarter-full bucket. According to the equation Fnet = m•v2 / R, force and mass are directly proportional. So the greater the mass, the greater the force.
25Example 3The diagram shows a 5.0-kilogram cart traveling clockwise in a horizontal circle of radius 2.0 meters at a constant speed of 4.0 meters per second. If the mass of the cart was doubled, the magnitude of the centripetal acceleration of the cart would bedoubledhalvedunchangedquadrupledac = v2 / R
26Example 4Two masses, A and B, move in circular paths as shown in the diagram. The centripetal acceleration of mass A, compared to that of mass B, isthe sametwice as greatone-half as greatfour times as greatF = m v2 / r
27Example 5A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.Given: = 900 kg; v = 10.0 m/s R = 25.0 mFind: a = ? Fnet = ?a = v2 / Ra = (10.0 m/s)2 / (25.0 m)a = (100 m2/s2) / (25.0 m)a = 4 m/s2Fnet = m • aFnet = (900 kg) • (4 m/s2)Fnet = 3600 N
28Example 6A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.Given: m = 95.0 kg; R = 12.0 m; Traveled 1/4 of the circumference in 2.1 sFind: v = ? a = ? Fnet = ?a = v2 / Ra = (8.97 m/s)2/(12.0 m)a = 6.71 m/s2v = d / tv = (1/4•2•π•12.0m) /(2.1s)v = 8.98 m/sFnet = m*aFnet = (95.0 kg)*(6.71 m/s2)Fnet = 637 N
29Example 7Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters.Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes 29.3 s).Find: FnetStep 1: find speed: v = (2 • π • R) / T = 6.22 m/s.Step 2: find the acceleration: a = v2 / R = (6.22 m/s)2/(2.90 m) = 13.3 m/s2Step 3: find net force: Fnet = m • a = (40 kg)(13.3 m/s/s) = 532 N.