Physics Beyond 2000 Chapter 3 Circular Motion

Physics Beyond 2000 Chapter 3 Circular Motion http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/ucm.html

Uniform Circular Motion The path is the circumference of a circle with constant radius r. The speed is a constant. The direction of motion changes with time. r v

Uniform Circular Motion The path is the circumference of a circle with constant radius r. The speed is a constant. The direction of motion changes with time. r

Uniform Circular Motion The path is the circumference of a circle with constant radius r. The speed is a constant. The direction of motion changes with time. v v v v r

Uniform Circular Motion Period T is the time needed to complete one cycle. Frequency f is the number of cycles completed in one second. r

Uniform Circular Motion Example 1

Uniform Circular Motion Angular displacement  –It is the angle, in radian, that the object turns.  r

Angular displacement Angular displacement  –It is the angle, in radian, that the object turns.  r s Length of arc s = r. 

Angular displacement After one complete cycle, angular displacement θ= 2π

Angular displacement Example 2

Angular speed Definition of average angular speed, ω av –Δθis the angular displacement –Δt is the time taken

Angular speed If we consider one complete cycle, Δθ= 2π and Δt = T then ω av = 2πf

Angular speed Example 3

Instantaneous angular speed

Example 4

Angular speed and linear speed When the object moves from A to B at linear speed v, Δs = r. Δθ  v = r. ω r Δs A B Δθ O X v

Angular speed and linear speed Example 5

Centripetal acceleration In a uniform circular motion, the velocity v changes in direction but not in magnitude. It requires an acceleration a to change the direction of the velocity but not the magnitude. The acceleration must be always perpendicular to the velocity. v a

Centripetal acceleration r A B Δθ O X vAvA vBvB In time Δt, the object moves from A to B. |v B | = |v A | = v

Centripetal acceleration r A B Δθ O X vAvA vBvB In time Δt, the object moves from A to B. Δθ Note that the triangle is an isosceles triangle. |v B | = |v A | = v

Centripetal acceleration r A B Δθ O X vAvA vBvB In time Δt, the object moves from A to B. Δθ |v B | = |v A | = v v. Δθ= a. Δt  a =

Centripetal acceleration r A B Δθ O X vAvA vBvB |v B | = |v A | = v

Centripetal acceleration r A B Δθ O X vAvA vBvB |v B | = |v A | = v The acceleration is pointing to the centre of the circle.

Centripetal acceleration The acceleration is pointing to the centre of the circle. The magnitude of the acceleration is or r. ω 2 r O X v a

Centripetal acceleration r O X v a In this motion, though the magnitude of the acceleration does not change, its direction changes with time. So the motion is of variable acceleration. r O X v a

Centripetal acceleration Example 6

Centripetal force Force produces acceleration. Centripetal force produces centripetal acceleration.

Centripetal force or The force F c is pointing to the centre of the circle. The force F c is perpendicular to the direction of the velocity. r O X v FcFc

Centripetal force or r O X v FcFc To keep the object moving in a circle of radius r and speed v, it is necessary to have a net force, the centripetal force, acting on the object.

Centripetal force Example 7

Centripetal Force: Example 7 The man is in circular motion. The net force on the man = F c. FcFc

Centripetal Force: Example 7 W – N = F c mg – N = F c N = mg - F c WN There are two forces on the man. N = normal contact force W = weight FcFc

Centripetal force If the provided force = then the object is kept in a uniform circular motion. r

Centripetal force If the provided force > then the object is circulating towards the centre. r

Centripetal force If the provided force < then the object is circulating away from the centre. r

Whirling a ball with a string in a horizontal circle Top View

Whirling a ball with a string in a horizontal circle Top View There is force F c acting on the ball along the string. There is force F acting at the centre along the string. FcFc F v

Whirling a ball with a string in a horizontal circle Top View The two forces F c and F are action and reaction pair. FcFc F v FcFc F v

Whirling a ball with a string in a horizontal circle Top View What happens if F c suddenly disappears? (e.g. the string breaks.) FcFc F v FcFc F v

Whirling a ball with a string in a horizontal circle What happens if F c suddenly disappears? (e.g. the string breaks.) v

Whirling a ball with a string in a horizontal circle What happens if F c suddenly disappears? (e.g. the string breaks.) It is moving away tangent to the circle.

Example 8 m=0.4kg M=0.5kg r = 0.5 m FcFc What is the source of the centripetal force? In equilibrium. http://www.dipmat.unict.it/vpl/ntnujava/circularMotion/circular3D_e.html In circular motion

Centripetal force or

Centripetal force F c  m F c  v 2 F c 

Uniform motion in a horizontal circle R  v

R  T mg The object is under two forces, the tension T and the weight mg.

Uniform motion in a horizontal circle R  T mg The net force on the object is the centripetal force because the object is moving in a circle. FcFc

Uniform motion in a horizontal circle R  T mg FcFc  T.cos  = mg and

Uniform motion in a horizontal circle R  T mg FcFc 

Uniform motion in a horizontal circle R  T mg FcFc  For a faster speed v, the angle θtends to increase.

Experiment To verify the equation for centripetal force or

To verify the equation for centripetal force Whirl the bob in a horizontal circle with string. The other end of the string is tied to some hanging weight. Maintain the hanging weight in equilibrium. bob hanging weight hollow plastic tube

To verify the equation for centripetal force L = length of the string in motion m = mass of the bob M = mass of the hanging weight ω= angular velocity of the bob θ= angle that the string makes with horizontal bob hanging weight L θ m M

To verify the equation for centripetal force T = tension on the string (tensions on both ends are equal if there is not any friction between the string and the tube.) Mg = weight of the hanging weight mg = weight of the bob bob hanging weight Mg T T mg θ

To verify the equation for centripetal force The hanging weight is in equilibrium. T = Mg ----------- (1) bob hanging weight Mg T T mg θ

To verify the equation for centripetal force The bob is in circular motion with angular velocity ω. F c = m.r. ω 2 ----------- (2) bob hanging weight Mg T T mg r L θ

To verify the equation for centripetal force bob hanging weight Mg T T mg r L FcFc θ The net force on the bob is equal to the centripetal force. (3)

To verify the equation for centripetal force bob hanging weight Mg T T mg r L FcFc θ Resolve the forces on the bob into vertical and horizontal components. F c = T.cosθ ------------- (4) and mg = T.sinθ------------ (5)

To verify the equation for centripetal force bob hanging weight Mg T T mg r L FcFc θ Also cosθ= (6)

To verify the equation for centripetal force bob hanging weight Mg T T mg r L FcFc θ From equations (1), (2), (4), (5) and (6), find ωin terms of L, m, M and g.

To verify the equation for centripetal force bob hanging weight Mg T T mg r L FcFc θ

Measure M and m before the experiment. bob hanging weight Mg T T mg r L FcFc θ

Measure ω during the experiment. bob hanging weight Mg T T mg r L FcFc θ ω = Number of revolution × 2π÷ time

Verify the following equation. bob hanging weight Mg T T mg r L FcFc θ

Problem: How to measure L? Refer to the textbook for the skill. bob hanging weight Mg T T mg r L FcFc θ

bob T mg r L FcFc θ Note that the bob must have a net force, the centripetal force, on it in order to keep it in a circular motion. As a matter of fact, the bob is not in equilibrium. It is in a motion with variable acceleration.

Leaning on a vertical cylinder Place an object on the inner wall of the cylinder. The cylinder starts to rotate about its axis. r

Leaning on a vertical cylinder As the cylinder rotates, the object performs a circular motion. At a certain angular velocity ω, the static friction may be sufficient to support the object on the wall without touching the ground. ω r

Leaning on a vertical cylinder There are 3 forces on the object. N = normal reaction W = weight of the object = mg f = static friction = μ s.N where μ s is the coefficient of static friction. N f W ω r

Leaning on a vertical cylinder As the object is in a circular motion, the net force must be the centripetal force. N = mrω 2 and μ s N ≧ mg Note that the static friction (f) cancels the weight (W). But the left hand side on the second equation is the limiting static friction which is the maximum friction. N f W ω r

Leaning on a vertical cylinder Solve the two equations. We have N f W ω r and

Rounding a Bend http://oldsci.eiu.edu/physics/DDavis/1150/05UCMGrav/Curve.html

Rounding a Bend r A car turns round a corner. It is a circular motion with a path of radius of curvature r.

Rounding a Bend r A car turns round a corner. It is a circular motion with a path of radius of curvature r. v

Rounding a Bend http://plabpc.csustan.edu/general/tutorials/CircularMotion/CentripetalAcceleration.htm

Rounding a Bend r It requires a centripetal force for the circular motion. v FcFc

Rounding a Bend How comes the centripetal force? r v FcFc It may come from the friction or the normal reaction.

Level Road without Banking r v FcFc F c comes from the static friction f s between tyres and the road. The speed v of the car must not exceed. where μ s is the coefficient of static friction.

Level Road without Banking r v FcFc v max = Note: v max is independent of the mass of the car.

Force on the passenger The passenger needs a centripetal force for turning round the corner with the car. The normal contact force from the car is the centripetal force.

Example 9 To find the coefficient of static friction. r v FcFc

Don’t rely on friction! When the road condition changes (e.g. on a rainy day), μ s becomes very small. Even a slow speed may exceed the safety limit. v max =

Banked Road Design a banked road which is inclined to the centre.

Banked Road Design a banked road which is inclined to the centre. R W θ The car is moving forward (into the plane) with velocity v and is turning left. The radius of curvature is r. r

Banked Road The centripetal force comes from the normal contact force R. R W FcFc θ r

Banked Road The centripetal force comes from the normal contact force R. Note that F c is the horizontal component of R. R W FcFc θ r

Banked Road In ideal case, friction is not necessary. The ideal banking angle is R W FcFc θ r

Example 10 Find the ideal banking angle of a road. The ideal speed is

Banked Road In non-ideal case, friction f is needed. Speed is too slow, less than the ideal speed. R W FcFc θ f r

Banked Road In non-ideal case, friction f is needed. Speed is too fast, more than the ideal speed. R W FcFc θ f r

Railway When there is a bend, the railway is banked. This avoids having lateral force on the rail.

Aircraft r FLFL W Back view of the car. The aircarft is moving forward (into the plane) with velocity v. When an airplane moves in a horizontal circular path in air, it must tilt about its own axis an angle θ. The horizontal component of the lift force F L is the centripetal force F c. θ

Aircraft When an airplane moves in a horizontal circular path in air, it must tilt about its own axis an angle θ.

Aircraft When an airplane moves in a horizontal circular path in air, it must tilt about its own axis an angle θ. The horizontal component of the lift force F L is the centripetal force F c. r FLFL W FcFc Note that F c is horizontal. θ

Aircraft When an airplane moves in a horizontal circular path in air, it must tilt about its own axis an angle θ. The horizontal component of the lift force F L is the centripetal force F c. r FLFL W FcFc θ

Example 11 Find the speed of the aircraft.

Bicycle on a Level Road When turning round a corner, it needs centripetal force. Like a car bending round a corner, the centripetal force comes from the static friction between the tyres and the road.

Bicycle on a Level Road Unlike a car, the bike inclines towards the centre to avoid toppling.  The bike is moving into the plane at speed v and is turning left. The radius of curvature is r. vertical horizontal r

Bicycle on a Level Road What is the angle of tilt  ?  The bike is moving into the plane at speed v and is turning left. The radius of curvature is r. vertical horizontal r

Bicycle on a Level Road Forces on the bike: weight W, normal contact force R and static friction f s.  vertical horizontal W R fsfs Note that W acts at the centre of mass G of the bike. h is the height of G from the ground. G h

Bicycle on a Level Road R = mg ------- (1) ----------- (2)  vertical horizontal W R fsfs G R balances W. f s is the centripetal force. h

Bicycle on a Level Road R = mg ------- (1) ----------- (2)  vertical horizontal W R fsfs G In order not to topple, the moment about G must be zero. About G, clockwise moment = anticlockwise moment h

Bicycle on a Level Road R = mg ------- (1) ----------- (2)  vertical horizontal W R fsfs G About G, clockwise moment = anticlockwise moment f s.h = R.a h

Tilt of a Car in Circular Motion Forces on the car: frictions f 1 and f 2, normal contact forces R 1 and R 2, weight W. The car is moving into the plane at speed v and is turning left. The radius of curvature is r. r

Tilt of a Car in Circular Motion Forces on the car: frictions f 1 and f 2, normal contact forces R 1 and R 2, weight W. r f1f1 f2f2 R1R1 R2R2 W acts at the centre of mass G of the car. W G

Tilt of a Car in Circular Motion Forces on the car: frictions f 1 and f 2, normal contact forces R 1 and R 2, weight W. r f1f1 f2f2 R1R1 R2R2 We are going to compare R 1 and R 2. W G

Tilt of a Car in Circular Motion Let 2L be the separation between the left and right tyres. r f1f1 f2f2 R1R1 R2R2 W G LL

Tilt of a Car in Circular Motion Let h be the height of the centre of mass G from the ground. r f1f1 f2f2 R1R1 R2R2 W G LL h

Tilt of a Car in Circular Motion Without toppling, the moment about G must be zero. About G, clockwise moments = anticlockwise moments r f1f1 f2f2 R1R1 R2R2 W G LL h

Tilt of a Car in Circular Motion r f1f1 f2f2 R1R1 R2R2 W G LL h So R 2 > R 1

Tilt of a Car in Circular Motion r f1f1 f2f2 R1R1 R2R2 W G As R 2 > R 1, the springs on the right are compressed more. The car tilts right when it turns left. http://www.sciencejoywagon.com/physicszone/lesson/03circ/centrif/centrif.htm

Uniform Motion in a Vertical Circle The path is the circumference of a vertical circle with constant radius r. The speed is v, a constant. The mass of the object is m.

Uniform Motion in a Vertical Circle The centripetal force is r FcFc v O How comes the centripetal force?

Uniform Motion in a Vertical Circle r FcFc v O The centripetal force comes from the tension T and the weight of the mass W or mg.

At the highest position T1T1 mg T 1 + mg = F c and r FcFc O v 

At the highest position T1T1 mg r FcFc O v

At the highest position T1T1 mg r FcFc O v What would happen if v 2 = ?

At the lowest position T2T2 mg T 2 - mg = F c and  O v FcFc Note that T 2 is always positive. r

At any other positions O v T3T3 r θ mg F There are three forces on the mass. T 3 is the tension from the rod, F is force from the rod and mg is the weight of the mass

At any other positions The net force is the centripetal force F c. O v FcFc r θ

At any other positions O v T3T3 r θ mg.cosθ θ Along the radial direction, F c = T 3 – mg.cosθ

At any other positions O v T3T3 r θ mg.cosθ θ

Uniform Motion in a Vertical Circle At the highest position, At the lowest position, At any other positions,

Non-uniform Motion in a Vertical Plane The motion of an object coasting along a vertical “ looping-the-loop”. Its speed would change at different positions. The principle is also applied to whirling mass with a string in a vertical plane.

Looping the loop Mass of the marble is m. Radius of the loop is r. The marble starts at the lowest position with speed v o. vovo O r

Looping the loop Note that there is change in kinetic energy and gravitational potential energy. Assume that energy is conserved. vovo O r vovo http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/ce.html

Looping the loop The speed v of the marble changes on the loop. The centripetal force changes on the loop. v O r v v

At the lowest position The speed v 1 of the marble is v o. The centripetal force F c comes from the normal contact force N 1 and the weight of the marble mg. O r vovo N1N1 mg

At the lowest position O r vovo N1N1 mg 

Below the centre O r v2v2 N2N2 mg  θ

Below the centre O r v2v2 N2N2 mg θ From conservation of energy, vovo h = r(1-cosθ)

Above the centre O r v3v3 N3N3 mg  ψ

Above the centre From conservation of energy, O r v3v3 N3N3 mg ψ h=r(1+cosψ) vovo

At the highest position v4v4 O r N4N4 mg 

At the highest position From conservation of energy, O r N4N4 mg h = 2r vovo v4v4

Completing the Circle For the marble to reach the highest position,  and 

Completing the Circle The marble cannot move up the loop and oscillates like a pendulum. vovo r

Completing the Circle The marble rises up to height more than r and is projected away. vovo r

Completing the Circle O r vovo

Whirling freely with a rod The ball moves and passes its loop with its own initial energy. vovo

Whirling freely with a rod A light rod would not be loosen. The light rod can provide tension or compression depending on the case. vovo

Whirling freely with a rod The minimum v o is for the marble to just reach the top. From conservation of energy, minimum v o = 2 min v o 2r v = 0 h

Whirling freely with a rod min v o 2r v θ h=r(1+cosθ) F mg θ

Changing from tension to compression min v o 2r v θoθo h=r(1+cosθ o ) mg θoθo When F = 0, the force changes from tension (F>0) to compression (F<0).

Changing from tension to compression min v o 2r v θoθo h=r(1+cosθ o ) mg θoθo Prove that θ o = 48.2 o when F = 0.

Example 12 Whirling a bucket of water in a vertical circle.

Example 12 Water does not flow out when the bucket is at the top position. v r mg

Centrifuge It is a device to separate solid or liquid particles of different densities by rotating them in a tube in a horizontal circle.

Centrifuge r axis of rotation ω ω is the angular velocity. r is the distance of the small portion of liquid from the axis of rotation.

Centrifuge r axis of rotation ω The small portion is in uniform circular motion. The centripetal force comes from the pressure difference ΔP. ΔP

Centrifuge r axis of rotation ω The small portion is replaced by another portion of smaller density. The centripetal force is not enough to support its uniform circular motion. ΔP

Centrifuge r axis of rotation ω As a result, this portion of smaller density moves towards the central axis. ΔP

Centrifuge r axis of rotation ω Portion of larger density moves away from the central axis. ΔP

Centrifuge Study p.51 and 52 for the mathematical deduction.

Physics Beyond 2000 Chapter 3 Circular Motion

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