TJC 7 Circular Motion

Student’s Copy Unit 6: MO Unit 7: Circular Motion Learning Outcomes Students should be able to: 7.1 Kinematics of uniform circular motion 7(a) express angular displacement in radians 7(b) show an understanding of and use the concept of angular velocity 7(c) recall and use v = r to solve problems 7.2 Centripetal acceleration 7(d) show an understanding of centripetal acceleration in the case of uniform motion in a circle, and qualitatively describe motion in a curved path (arc) as due to a resultant force that is both perpendicular to the motion and centripetal in direction 7(e) recall and use centripetal acceleration a = r2 and a = 2v r to solve problems 7(f) recall and use F = mr2 and F = 2mv r to solve problems.

2025 Temasek Junior College 2 1 Introduction We can find many examples of circular motion around us, e.g. bicycles and cars going round a corner and the Moon orbiting around the Earth. In the first part of this topic, we will describe the motion of objects moving in a circular path; in the subsequent parts, we will consider what causes motion to be circular. 2 Definition of Physical Quantities 2.1 Angular Displacement,  LO(a) In terms of the length of arc s and the radius of the circle r, angular displacement θ is expressed as       = s r  Angular displacement is measured in radian (rad).  For one complete revolution, θ = 2r r = 2 rad (i.e. 2 rad = 360o)  To convert θ in degrees to radians: θ(rad) 2π = θ(°) 360° Note: The radian is physically dimensionless as it is the ratio of two lengths. 2.2 Angular Velocity,  LO(b),(c) i.e.  = dθ dt unit of ω: rad s-1 Consider an object moving with constant speed v in a circular path of radius r. In time t the object moves along an arc of length s and sweeps out an angle .   = ∆θ ∆t   = ∆s r⁄ ∆t (since s = r)   = 1 r ∆s ∆t  = v r (since v = ∆s ∆t) Hence v = r One radian is defined as the angle subtended at the centre of a circle by an arc equal in length to the radius of the circle. Angular velocity is defined as the rate of change of angular displacement s  r

2025 Temasek Junior College 3 A B C 12 6 9 3 For uniform circular motion, since v has the same magnitude throughout the motion, angular velocity is a constant (equal angle is swept out in equal time intervals). If the time taken to complete one revolution (i.e. to turn through an angle of 2 radians) is T, then T is known as the period. Note: Angular velocity is a vector quantity. Example 1 (a) Which point A, B or C on the minute hand of a clock is moving with the greatest speed? (b) Calculate the angular velocity of the minute hand. Example 2 A model car moves round a circular track of radius 0.30 m at 2.0 revolutions per second. Calculate (a) the period T, (b) the angular speed , (c) the linear speed v of the car. (a)  1 0.50 s2.0T (b) 122 12.6 rad s0.50T       (c) v = r = 0.30  12.6 = 3.8 m s-1  = 2 T (a) Each point on the minute hand moves with same angular velocity,. Since v = r  point C, being farthest from the centre, has the greatest linear speed. (b) 3 -1rad s                                22 1.75 10 60 60T      

2025 Temasek Junior College 4 3 Uniform Motion in a Circle 3.1 Centripetal Acceleration LO (d),(e),(f) In a uniform circular motion, although the linear speed remains unchanged, the velocity is always changing (since the direction changes with time). So, the motion is an accelerated one and by Newton’s second law, a resultant force must be acting on it. Alternatively, you can view it as a body moving along a circular path requires a resultant force to act on it. Otherwise it would move in a straight line in accordance with Newton’s first law. In the above diagram, consider a very short interval of time so that points A and B are close. As the particle moves from A to B, its velocity changes from vi to vf. The change in velocity v is directed toward point C, the centre of the circle. Hence the acceleration is also directed towards the centre of the circle. This acceleration is known as centripetal acceleration and is given by a = 2v r = r2 3.2 Centripetal Force LO (f) Since F = ma, there must be a net force directed towards the centre of the circle to provide the acceleration. The force causing the circular motion is known as the centripetal force and is given by F = 2mv r = mr2 Note: 1. For an object to move in a circular path, a physical force must be present to provide for the centripetal force. The centripetal force is just a resultant force. As such, the centripetal force should not be drawn as an additional force in force diagrams. Δv = vf – vi = vf + (−vi) vi vf A B C Δv –vi vf

2025 Temasek Junior College 5 Hence for the above ball swinging in a horizontal circle, only t he tension T in the string providing the centripetal force is drawn, not the resultant centripetal force. If the string were to break, there would be no centripetal force and the ball would fly off in a direction tangent to the circular path. 2. Since the centripetal force acts at right angles to the motion, the centripetal force cannot do any work. (Recall: work done by a force is defined as the product of the force and displacement in the direction of the force). Hence the centripetal force does not change the K.E. or speed of the object. Example 3 Conical pendulum A mass of 0.50 kg is made to describe a circular path in a fixed horizontal plane. The string of length 25 cm makes an angle of 30o to the vertical. Calculate (a) the tension in the string, (b) the linear speed of the mass, and (c) the period of oscillation of the mass. (c) (b) T sinθ = mv2 r = mv2 Lsin𝜃 5.7 sin 30o = 0.50v2 (0.25)sin30o v = 0.84 m s-1 (a) Tcos θ = mg T cos30o = 0.50  9.81 T = 5.7 N r  T mg     2 2 0.25sin30 0.93 s0.84 o dist rT speed v direction of motion when string breaks v T tension T of string on ball provides centripetal force T Conical pendulum – a mass on a string oscillating in a circular path.

2025 Temasek Junior College 6 Example 4 Car turning around a bend A 1200 kg car is turning a bend at 8.0 m s-1, and it travels along an arc of a circle in the process. The radius of the circle is 9.0 m. (a) Calculate the horizontal force exerted by the road on the tyres that is required to hold the car in the circular path. (b) If the maximum frictional force between the road and the wheels is only 4.8  103 N, state and explain what will happen. (c) Calculate the maximum speed for safe turning. f is the sideways frictional force by road on tyres providing the centripetal force FC. (a) Fc = mv2 r = 1200  8.02/9.0 = 8.5  103 N (b) The maximum frictional force of 4.8  103 N is less than the centripetal force 8.5  103 N required to turn around the bend. Hence the car will skid outwards. (c) For safe turning, fmax = FCmax 4.8  103 = mv2 r = (1200)v2 9.0 vmax = 6.0 m s–1  centre of circle f N mg controlling a skid

2025 Temasek Junior College 7 3.2.1 Providing centripetal force using banking A banked turn is one in which the vehicle banks or inclines towards the inside of the turn. It is a common manoeuvre that increases the availability of forces that provide for the centripetal force required for a turn. Case Study 1: Airplane turning in a horizontal plane