Motion in a Circle JPJC Notes

JURONG PIONEER JUNIOR COLLEGE 9749 H2 PHYSICS MOTION IN A CIRCLE Content (I) Kinematics of uniform circular motion (II) Centripetal acceleration (III) Centripetal force Learning Outcomes Candidates should be able to: (a) express angular displacement in radians. (b) show an understanding of and use the concept of angular velocity to solve problems. (c) recall and use vr = to solve problems. (d) describe qualitatively motion in a curved path due to a perpendicular force, and understand the centripetal acceleration in the case of uniform motion in a circle. (e) recall and use centripetal acceleration 2ar = , and 2va r= to solve problems. (f) recall and use centripetal force 2F mr = , and 2mvF r= to solve problems.

2 Introduction ▪ In the previous topic on Kinematics, we learnt about uniformly accelerated motion. An object accelerating along a straight line experiences a net force that acts along its direction of motion. However, if a constant net force acts at an angle to the direction of motion at any instant, the object moves in a curved path (e.g. projectile motion). ▪ In this topic, we will study the circular motion of objects in which the acceleration is not uniform. In particular, we will focus on uniform circular motion in which an object travels at a constant speed. 1 Kinematics of uniform circular motion (a) Candidates should be able to express angular displacement in radians. 1.1 Angular displacement ▪ Consider an object movin g from A to B in a circle with uniform speed v round a fixed point O as centre. Fig. 1 Object moving in uniform circular motion from A to B ▪ The angle  swept through by the radius is known as the angular displacement. ▪ It is defined by the equation s r = where s is the arc length AB and r is the radius of the circle. ▪ The angular displacement  is measured in radians. One radian (rad) is defined as the angle subtended at the centre of a circle by an arc equal in length to the radius. r v v O A B r

3 (b) Candidates should be able to show an understanding of and use the concept of angular velocity to solve problems. 1.2 Angular velocity ▪ Angular velocity  is the rate of change of angular displacement. d dt  = It is measured in rad s−1. ▪ For an object moving with constant angular velocity, t  = . Example 1 A boy seated 2.0 m away from the centre of a merry -go-round completes one -sixth of a revolution in 3.0 s. Calculate (a) his angular displacement  from his starting position, (b) the distance he travels from his starting position, (c) his angular velocity. Solution: (a) For one revolution,  = 2 rad; for one-sixth of a revolution, angular displacement  1 26 = = 1.047 1.0= rad (b) The distance travelled is the arc length s of the circle. s r= 2.0 1.047= 2.1= m (c) Angular velocity 1.047 3.0t  == 0.35= rad s−1 (c) Candidates should be able to recall and use vr = to solve problems. 1.3 Tangential velocity ▪ The tangential velocity v is the instantaneous velocity of the object along its circular path. The direction is therefore tangential to the circular path. ▪ Referring to Fig . 1, recall that if s is the length of the arc AB, then sr = . Differentiating the equation with respect to time t, we have ()ds d r d rdt dt dt ==

4 Hence, the following relationship is obtained. vr = ▪ This relationship shows that the larger the radius r, the larger the tangential velocity v for a constant angular velocity  . 1.4 Period and frequency ▪ The period T of a circular motion is defined as the time taken for an object to complete one revolution. In time T, the object moves through an angle of 2 rad. Therefore, 2 T  = 2T  = ▪ The frequency f of a circular motion is defined as the number of revolutions completed per unit time. It is related to the period by 1f T= Also,  = 2f Example 2 A toy car moves round a circular track of radius 0.30 m at 2.0 revolutions per second. Calculate its (a) period T, (b) angular velocity , (c) tangential velocity v. Solution: (a) Period 1T f= 1 2.0= 0.50= s (b) Angular velocity 2 T  = 2 0.50 = = 12.566 13= rad s−1 (c) Tangential velocity vr = 0.30 12.566= 3.8= m s−1

5 2 Centripetal acceleration (d) Candidates should be able to describe qualitatively motion in a curved path due to a perpendicular force, and understand the centripetal acceleration in the case of uniform motion in a circle. 2.1 Centripetal acceleration ▪ The velocity of an object in circular motion is always changing even when the speed is constant, since the direction is always changing; velocity is a vector, so direction must be taken into account. ▪ This means that even when a n object is moving with constant speed in a circular path, it is still accelerating. ▪ This acceleration cannot have a component in the direction of motion of the object, for if it had, it would either increase or decrease the speed of the object. ▪ Hence, this acceleration must be perpendicular to the direction of motion of the object, and is directed towards the centre of the circle. (e) Candidates should be able to recall and use centripetal acceleration 2ar = , and 2va r= to solve problems. 2.2 Derivation of centripetal acceleration ▪ Consider an object moving from A to B in a circle with uniform speed v. It moves through an angle of  in time t . Fig. 2 Object moving in uniform circular motion from A to B r O A B r

6 ▪ The change in velocity of the object is given as BAv v v = − , where Av and Bv are the velocities of the object at A and B respectively. BAv v v = − ( )BAv v v = + − Note: ABv v v== ▪ Consider the angle  to be very small (when t is very small). Therefore, vv    , where v is directed towards the centre of the circle. ▪ Since acceleration is defined as the rate of change of velocity, va t =  va t =  av = ▪ The acceleration is in the same direction as the change of velocity, which is towards the centre of the circle. It is known as the centripetal acceleration, ca and can be written as 2 2 c va v r r= = = Example 3 The Moon orbits around the Earth with a speed of 1020 m s−1 and takes 27.3 days for a complete revolution. Calculate (a) the acceleration of the Moon, (b) the distance between the centres of the Moon and Earth. Solution: (a) Centripetal acceleration, cav = 2v T =  21020 27.3 24 3600 =   32.72 10 −= m s−2 (b) Let r be the distance between the centres of the Moon and Earth (which is the radius of the orbit of the Moon, assumed circular). Since vr = , 21020 27.3 24 3600r =  83.83 10r = m

7 2.3 Summary of terms ▪ Fig. 3 gives a summary of the terms used for circular motion and the relationship between them. tangential angular relationship displacement s  sr = velocity v  vr = Fig. 3 Summary of physical quantities used for circular motion ▪ Centripetal acceleration, 2 2 c var r== . 3 Centripetal force (d) Candidates should be able to describe qualitatively motion in a curved path due to a perpendicular force, and understand the centripetal acceleration in the case of uniform motion in a circle. 3.1 Centripetal force ▪ Applying Newton’s second law of motion, it can be deduced that i f an object is moving along a circular path, there must be a resultant force acting on it since there is acceleration. ▪ The direction of this resultant force is in the sa