DHS 06 Motion in a Circle (Lecture Notes & Tutorial)

Dunman High School (Senior High Physics Department) 9749 Physics (2024) Topic 6: Motion in a Circle Topic 6 - Student Notes - Page 1 Guiding Questions  How do we describe the motion of an object moving in a circular path?  What causes an object to move in a circular path? Content  Kinematics of uniform circular motion  Centripetal acceleration  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 v = rω 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 a = rω2, and a = v2/r to solve problems (f) recall and use centripetal force F = mrω2, and F = mv2/r to solve problems. 9749 H2 Physics Topic 6 Motion in a Circle Year 5 (2024) DUNMAN HIGH SCHOOL

Dunman High School (Senior High Physics Department) 9749 Physics (2024) Topic 6: Motion in a Circle Topic 6 - Student Notes - Page 2 6.1 Circular Motion: An Introduction (a) express angular displacement in radians. So far, we have discussed the motion of an object in a straight line (linear motion) and a parabolic path (projectile motion). However, in real life, there are many cases of objects moving in an arc of a circle or a circular path. For example, we can tie a ball to a string and swing it in a horizontal circle. Suppose the radius of this circle is r and the ball travels a distance (arc length) of s, its angular displacement will be θ. Angle θ in radians is the ratio of the distance s along a circular arc subtended by θ divided by the radius r. s r θ= If s = r, then θ =1 radian. Since circumference of circle ,diameter of circle  the circumference of a circle = 2πr. So for one revolution 2 .2 2 s r r r     θ= o o o 2 radian 360 360Thus, 1 radian 57.32      Conversion to degree can be done using: o (rad) 180      Translate degrees to radians: degrees180 180 Translate radians to degrees: r adians      One radian is

Dunman High School (Senior High Physics Department) 9749 Physics (2024) Topic 6: Motion in a Circle Topic 6 - Student Notes - Page 3 6.1.1 Angular velocity ω (b) show an understanding of and use the concept of angular velocity to solve problems. Angular velocity, ω, is the rate of change of angular displacement. = d dt  Its unit is rad s−1. When the angular velocity is constant, t   angle moved through in time time taken t the body is then said to be undergoing uniform circular motion, i.e. the motion of an object moving in circular path at constant speed. For our syllabus, we are mainly interested in objects moving in uniform circular motion. If a body undergoing uniform circular motion takes T seconds to complete 1 full revolution, i.e., its period is T seconds, where f represents the frequency of the circular motion. 6.1.2 Relationship between angular velocity ω and linear speed v for uniform circular motion (c) recall and use v = rω to solve problems. From s r , the linear speed of an object (e.g. P , as shown in the diagram) can be determined via calculus. ds d drrdt dt dt    If r is a constant, ds d rdt dt  Therefore, This relationship is only valid when ω is measured in rad s−1. Note that the direction of the (linear) velocity is tangential to the circular path. For an object rotating about an axis, every point on the object has the same angular velocity. The tangential speed of any point is proportional to its distance from the axis of rotation.

Dunman High School (Senior High Physics Department) 9749 Physics (2024) Topic 6: Motion in a Circle Topic 6 - Student Notes - Page 4 Question 6.1 Find the angular velocity of the second, minute and hour hands of a clock. If the second hand, minute hand and the hour hand is 5.0 cm, 5.0 cm and 4.0 cm long respectively, find the linear speed of the tip of each hand. 1 3 1 2 2Using , for second hand, 0.105 rad s 2 for minute hand, 1.75 10 rad s 2 for hour hand, T                4 11.45 10 rad s          3 1 3 5 1 Using , for second hand, 0 .105 5.25 10 m s for minute hand, 1.75 10 8.75 10 m s for hour hand, v r v v v                  4 6 1 1.45 10 5.80 10 m s     Question 6.2 A Blu-ray disc (BD) rotating inside a player must maintain a constant linear speed of 4.92 m s−1 in order for the information on it to be read correctly. The laser in the Blu-ray player starts at the inside of the disc and moves outwards. (a) State and explain what must happen to the angular velocity of the disc as the BD is played. (b) Calculate the radius at which the BD will have an angular velocity of 23.1 revolutions per second when the linear speed is 4.92 m s−1. Answers: (a) The angular velocity of the disc must ……………………. when the laser is moving outwards to read the info. Sincev r  , when r increases, must ……………….. to keep v constant. (b)

Dunman High School (Senior High Physics Department) 9749 Physics (2024) Topic 6: Motion in a Circle Topic 6 - Student Notes - Page 5 6.2 Analysis of uniform circular motion 6.2.1 Centripetal acceleration in uniform circular motion (e) recall and use centripetal acceleration a = rω2, and a = v2/r to solve problems. A body is moving round a circle radius r with constant angular velocity ω. It moves from point A to point B in a brief time of δt. Its velocity changes from vA to vB in this time (change in direction but not magnitude). In time δt, the body moves round the circle a distance arc AB of length r δθ. Its speed .v r r t    The vector diagram shows that the change in velocity in time δt, δv = v δθ (if δθ is small) The average acceleration of the body between A and B (rate of change of velocity) = v v t t     In the limit when t approaches zero, 0 lim t d t dt      , The acceleration of the body at any point is As δθ becomes infinitesimally small, δv is perpendicular to vA. Since δv is in the same direction as the acceleration, the direction of acceleration of the body undergoing uniform circular motion is towards the centre of the circular path. This acceleration is called the centripetal acceleration. Centripetal acceleration is one which is always perpendicular to the velocity and always acts towards the centre of the circular motion. This centripetal acceleration can also be expressed as change in velocity, δv =

Dunman High School (Senior High Physics Department) 9749 Physics (2024) Topic 6: Motion in a Circle Topic 6 - Student Notes - Page 6 6.2.2 Centripetal force in uniform circular motion (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. Recall: An object at rest will remain at rest and an object in motion will remain in motion at constant velocity in the absence of an external resultant force. Consider an object moving in uniform circular motion at constant angular velocity, , from A to B. If no external resultant force acts on the object, by Newton’s 1st Law (N1L), the object will travel from A to C with a constant velocity, v. However, since the object travels along the curved path from A to B, this implies that an externa