Circular motion A particle P travels in circular path. It performs circular motion about O with radius r. s P w A If P travels in constant speed, the circular motion is known as uniform.

Mathematics – angles 1 cm q Arc length = circumference x q /360o 1 = 2 p x q / 360o q =180o / p ≈ 57o

Angle in radian (another unit for angle) q r s If s = r, then q is 1 radian. 1 radian = 180o / p 180o = p radian 360o = 2 p radian

Relation between arc length s, angle at centre q and radius r is q = s/r Relation between arc length s, angle at centre q and radius r is s = rq

Describing circular motion Consider a body moving uniformly from A to B in time t so that it rotates through an angle q . Angular displacement (q / rad) Anticlockwise as positive direction Angular velocity (w / rad s-1) = Angular displacement / time * s B w A Speed of the body By s = rq

Speed and angular speed Consider carts A and B in a mechanical game. Do they have the same angular speed? Which of them moves faster? 2 m A 1 m B w = 3 rad s-1

Describing circular motion Period of the motion s B w A = distance / speed Period of the motion = angular displacement / angular speed

Three useful expressions (1) s = rq (2) v = rw (3) 180o = p radians

Example 1 Find the angular velocity in radian s-1if a motor makes 3000 revolutions per minute. Solution: 1 revolution = 360o = 2p radians. Angular velocity = 3000 x 2p / 60 = 314 rad s-1

Rotational motion and Translational motion q: angular displacement w: angular velocity a: angular acceleration Equations of motion for uniform acceleration: w1 = w0 + at q = w0t + ½ at2 w12 – w02 = 2aq w q v = u + at s = ut + ½ at2 v2 –u2 = 2as

Example A turntable is rotating at 1 rev / s initially. A motor is turned on such that the turntable rotates at an angular acceleration of 0.3 radian s-2. Find the angular displacement covered by the turntable for 5 s just after the motor is turned on. Solution: w0 = 1 rev / s = 2p radian s-1 (∵360o = 2p radians) a = 0.3 radian s-2 t = 5 s By q = w0t + ½ at2 q = (2p)(5) + ½ (0.3)(5)2 = 35.2 radians

Centripetal acceleration A body which travels equal distances in equal times along a circular path has constant speed but not constant velocity. Since the direction of the velocity changes from time to time, the body has acceleration.

Centripetal acceleration (Deriving a = v2/r) Consider a body moving with constant speed v in a circle of radius r. It travels from A to B in a short interval of time dt. vB vA dv vA B w vB dq dq O A r But v = rw, Therefore, Change in velocity = vB – vA = (– vA) + vB = dv Since d t is small, dq will also be small, so dv = vdq . Acceleration Since dq is small, dv will be perpendicular to vA and points towards the centre. Hence, the direction of acceleration is towards the centre. i.e. centripetal.

Centripetal force Since a body moving in a circle (or a circular arc) is accelerating, it follows from Newton’s first law of motion that there must be a force acting on it to cause the acceleration. The direction of this force is also towards the centre, therefore this force is called centripetal force. By Newton’s second law, the magnitude of centripetal force is Since the centripetal force  displacement, NO work is done by the centripetal force ⇒ K.E. of the body in uniform circular motion remains unchanged.

A stone of mass 2 kg is tied by a string and moving in a horizontal circular path of radius 0.5 m. Find the tension in the string if the speed of the stone is 4 ms-1. Solution: Tension =Centripetal force = mv2/r = (2)(4)2/0.5 = 64 N Smooth table

Examples of circular motion A bob is tied to a string and whirled wound in the horizontal circle. In vertical direction: w T mg q In horizontal direction:

Rounding a bend Case 1: Without banking r When a car of mass m moving with speed v goes around a circular bend with radius of curvature r, the centripetal force f r The frictional force between the tires and the road provides the centripetal force. If limiting friction < , the car slips on the road and it may cause an accident.

Limiting friction is attained to achieve maximum speed. The coefficients of friction between the tires and the road in rainy days and in sunny days are 0.4 and 0.6 respectively. If a car goes around a circular bend without banking of radius of curvature 50 m, find the maximum speeds of the car without slipping in rainy days and sunny days. Solution: Limiting friction is attained to achieve maximum speed. By Maximum speed without slipping = mrg In rainy days, Maximum speed without slipping = ms-1 (50.9 km h-1) In sunny days, Maximum speed without slipping = ms-1 (62.4 km h-1)

In vertical direction: Rounding a bend In order to travel round a bend with a higher safety speed, the road is designed banked. The centripetal force does not rely on friction, but provided by the horizontal component of the normal reaction. Case 2 with banking mg r R q In vertical direction: In horizontal direction: The safety speed of rounding the bank is which can be increased by increasing the degree of banking of the road.

The figure above shows a car moving round a corner with a radius of 8 m on a banked road of inclination 20o. At what speed would there be no friction acting on the car along OA? Solution: In vertical direction: In horizontal direction: ms-1 . Note: If the car travels at a higher speed say 6 ms-1 round the above corner, some frictional force would act on the car to prevent the car from slipping upward along the road provided that the limiting friction is not exceeded i.e.

Aircraft turning in flight mg L L cos q q L sin q L mg straight level flight turning in flight In straight level flight, the wings provide a lifting force L that balances its weight mg. i.e. L = mg. To make a turn at v with radius r, the flight banks and the horizontal component of the lifting force (L sin q) provides the centripetal force for circular motion.

Aircraft turning in flight L cos q q To make a turn at v with radius r, the flight banks and the horizontal component of the lifting force (L sin q) provides the centripetal force for circular motion. L sin q mg turning in flight The weight of the aircraft is supported by the vertical component of L. i.e.

Find the angle of inclination of the wings of an aircraft which is traveling in a circular path of radius 2000 m at a speed of 360 km h-1. mg L Lcos q Lsin q q Solution: Speed of the aircraft = 360 /3.6 ms-1 = 100 ms-1 Resolve horizontally, L sin q = mv2 / r --- (1) Resolve vertically, L cos q = mg --- (2) (1) / (2): tan q = v2 / rg tan q = 1002 / (2000 x 10) q = 26.6o

Cyclist rounding a corner The frictional force provides the centripetal force. By the law of friction, the maximum speed of the cyclist without slipping is given by C.G. mg R f r However, f also has a moment about C.G which tends to turn the rider outwards.

The rider must lean inwards so that the moment of f is counterbalanced by the moment of R about C.G.. C.G. mg R f r q h a For no overturning of the cyclist, C.G. Take moment about C.G..: In vertical direction: Sub (2) and (3) into (1): In horizontal direction: ∴ For given v and r, the cyclist must bend at the correct angle in order not to overturn.

The rotor f R mg A mechanical game in amusement parks. It consists of an upright drum, inside which passengers stand with their backs against the wall. The drum spins at increasing speed about its central vertical axis. Hence, the centripetal force required for circular motion also increases. The centripetal force is provided by the normal reaction from the wall. i.e. mg f R Therefore, the normal reaction increases as the spinning speed increases. As normal reaction increases, the weight of the passenger can be balanced by the frictional force. i.e. When the floor is pulled downwards, the passenger will not fall but remains stuck against the wall of the rotor.

It is given that the radius of the drum is 2 m and the coefficient of static friction between clothing and the wall is 0.4 m. Find the minimum speed v of the passenger before the floor is pulled downwards? mg f R Solution: In horizontal direction: In vertical direction: By the law of friction: Sub. (1) and (2) into (3): The minimum speed required is 7.07 ms-1

Looping the loop r u v R mg C B D A At C, the centripetal force is provided by the normal reaction and the weight of the object. r u v R mg C B D If the object does not leave the track, R  0. The minimum speed at C to just complete the loop is given by A By conservation of energy, the minimum speed at A for the object to complete the loop is given by Sub. (1) into (2):

r u v R mg C B D Notice that: If the ball passes point A with speed , the ball will fall from the track. , the ball will complete the loop If the ball passes point A with speed and the remaining centripetal force is provided by the normal reaction.

Weightlessness Weightlessness means the weight of an object is equal to zero. This happens at a place where there is no gravitational field (g = 0). We are aware of our weight because the ground exerts an upward push (normal reaction) on us. If our feet are completely unsupported, for example, in a free falling lift, we experience the sensation of ‘weightlessness’. An astronaut orbiting the earth in a space vehicle with its rocket motors off also experiences the sensation of ‘weightlessness’.

An astronaut orbiting the earth An astronaut orbiting the earth in a space vehicle with its rocket motors off also experiences the sensation of ‘weightlessness’. Why? The weight of the astronaut provides the centripetal force and the walls of the vehicle exert no force on him. v mg r i.e. Note: g < 10 ms-2 because the astronaut is now far away from the surface of the earth.

Experience the sensation of ‘weightlessness’ Summary Weightlessness Experience the sensation of ‘weightlessness’ W = 0 W > 0 but R = 0 At a place where there is no gravitational field (g = 0). In a free falling lift. In a space vehicle with its rockets off orbiting the earth. Normal reaction R = 0

Centrifuge (離心機) Centrifuges separate solids suspended in liquids, or liquids of different densities. The mixture is in a tube, and when it is rotated at high speed in a horizontal circle, the less dense matter moves towards the centre of rotation. On stopping the rotation, the tube returns to the vertical position with the less dense matter at the top. Cream is separated from milk in this way.

Working principle of a centrifuge Consider the part of liquid between A and B inside the tube. The pressure at B is greater than that at A. This provides the necessary centripetal force acting inwards. For this part of liquid, the force due to pressure difference supplies exactly the centripetal force required. If this part of liquid is replaced by matter of small density or mass, the centripetal force is too large and the matter is pushed inwards. On stopping the rotation, the tube returns to the vertical position with the less dense matter at the top. pA pB B A pB > pA

H.W. Chapter 2 (1, 2, 3, 5, 6(a)(b)(c) ) Due date: 27/11 Day 1 1. A particle moves in a semicircular path AB of radius 5.0 m with constant speed 11 ms-1. Calculate (a) the time taken to travel from A to B (p = 22/7) (b) the average velocity, (c) average acceleration A 5 m 5 m B 11 ms-1

(a) its angular velocity in rad s-1, 2 A turntable of a record player makes 33 revolutions per minutes. Calculate (a) its angular velocity in rad s-1, (b) the linear velocity of a point 0.12 m from the centre. 33 rev / min v 180o = p radians v= rw

3. What is meant by a centripetal force. Why does 3 What is meant by a centripetal force? Why does such a force do no work in a circular orbit? (a) An object of mass 0.5 kg on the end of a string is whirled round a horizontal circle of radius 2.0 m with a constant speed of 10 ms-1. Find it angular velocity and the tension in the string. (b) If the same object is now whirled in a vertical circle of the same radius with the same speed, what are the maximum and minimum tensions in the string? A T + mg = mv2/r T – mg = mv2/r T = mv2/r B

Referring to the above diagram, find the normal reaction acting on the ball at A, B and C if the speed of the ball at A is C vc R When the ball is at A, mg vB Centripetal force: RA - mg = mu2/r RA = mg + m(6gr)/r = 7mg RB D B r mg RA u When the ball is at B, mg By conservation of energy, ½ mu2 = ½ mvB2 + mgr vB2 = u2 – 2gr = 6gr – 2gr = 4gr Centripetal force = RB = mvB2/r RB = m(4gr)/r = 4mg A When the ball is at C, By conservation of energy, ½ mu2 = ½ mvc2 + mg(2r) vc2 = u2 – 4gr = 6gr – 4gr = 2gr Centripetal force: RC + mg = mvB2/r RC = m(2gr)/r – mg = mg