Circular Motion Notes PDF

CIRCULAR MOTION This refer to the motion of an object in a circle about a fixed point (centre) Terms used  Angular velocity (𝝎) This is defined as the change of angle for a body moving in a circular path about a fixed point. OR It is defined as the rate of change of an angle of rotation for a...

CIRCULAR MOTION This refer to the motion of an object in a circle about a fixed point (centre) Terms used  Angular velocity (𝝎) This is defined as the change of angle for a body moving in a circular path about a fixed point. OR It is defined as the rate of change of an angle of rotation for a body moving in a circular path about the centre. Suppose a body moves from point A to point B, so that the radius OA moves through an angle 𝜃, then 𝑣𝐵 𝐵 𝑣𝐴 𝑠 𝑂 𝜃 𝐴  Period (Period), T This is the time taken by an object to make one complete cycle (revolution). The S.I unit is second. 1 𝑟𝑒𝑣 = 2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 𝜽 𝜽 𝐹𝑟𝑜𝑚, 𝝎 = ⇒𝒕= 𝒕 𝝎 𝑤ℎ𝑒𝑛 𝜃 = 2𝜋, 𝑡 = 𝑇 𝟐𝝅 ∴𝑻= 𝝎  Frequency, 𝒇 This is the number of complete cycles/revolutions made in one second. The S.I unit is Hertz (𝐻𝑧) 1 𝑓=𝑇 𝟐𝝅 𝐹𝑟𝑜𝑚 𝑻 = 𝝎 , 𝝎 ⇒ 𝒇 = 𝟐𝝅 𝒐𝒓 𝝎 = 𝟐𝝅𝒇  Linear speed, 𝑽 Speed is the rate of change of distance with time. An object moving in a circular motion moves with a constant speed but with changing velocity because of continuous change in direction. Consider a body moving in a circle of radius 𝒓 with uniform speed 𝒗 𝑣𝐵 𝐵 𝑣𝐴 𝑠 𝑂 𝜃 ` 𝐴 Suppose the body moves from point A to point B in time 𝒕 through an angle 𝜽. The angle 𝜃 is called the angular displacement. 𝜃 Arc length, 𝑠 = 2𝜋 × 2𝜋𝑟 = 𝑟𝜃 1  𝜃 𝜃 Angular velocity, 𝜔 = 𝑡 ⇒𝑡=𝜔 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒,𝑠 Speed,𝑣 = 𝑡𝑖𝑚𝑒,𝑡 𝜃 𝑣 = 𝑟𝜃 ÷ 𝜔 𝒗 = 𝝎𝒓  Acceleration of a body moving in a circular path Centripetal acceleration is defined as the rate of change of velocity for an object moving in a circular path and it is directed towards the centre of the circular path. Derivation of expression for centripetal acceleration Consider a body moving in a circle of radius 𝒓 with uniform speed, 𝒗. Let it move from 𝐴 to 𝐵 in time ∆𝑡 as shown in figure (i) 𝑣𝐵 𝐵 𝑣𝐵 −𝑣𝐴 𝑅 𝑃 𝜃 𝑣𝐴 𝑂 ∆𝜃 m 𝑣𝐵 −𝑣𝐴 𝐴 ∆𝜃 𝑄 (i) (ii) 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝒗𝑩 − 𝒗𝑨 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎 = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = 𝑡 ⃗⃗⃗⃗⃗ = 𝒗𝑩 − 𝒗𝑨 In figure (ii) 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑃𝑅 ⃗⃗⃗⃗⃗ approximates to an arc where |𝑣𝐴 | = |𝑣𝐵 | = 𝑣 is the radius. If ∆𝑡 is small, then ∆𝜃 is also small and 𝑃𝑅 ⃗⃗⃗⃗⃗ | = 𝑣∆𝜃 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ⇒ 𝐴𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ, |𝑃𝑅 ∴ 𝒗𝑩 − 𝒗𝑨 = 𝑣∆𝜃 𝑣∆𝜃 ∆𝜃 ⇒𝑎= 𝑡 but 𝑡 =𝜔 𝒂 = 𝒗𝝎 𝑣 From 𝑣 = 𝜔𝑟 ⇒ 𝜔 = 𝑟 𝒗𝟐 𝟐 ∴𝒂= Or 𝒂=𝝎 𝒓 𝒓 Alternative Consider a body of mass 𝒎 moving around a circular path of radius 𝒓 with uniform angular velocity 𝝎 and linear speed 𝒗. If initially the body is at point 𝐴 moving with velocity 𝒗𝑨 and after a small time interval 𝒕, the body is at point B where its velocity is 𝒗𝑩 with the radius having moved an angle, ∆𝜽 𝑣𝐵 Velocity at 𝐴, parallel to 𝑂𝐴, = 0 Velocity at 𝐵,perpendicular to 𝑂𝐴, = 𝑣𝑐𝑜𝑠∆𝜃 Velocity at 𝐴, perpendicular to 𝑂𝐴, = 𝑣 ⇒ 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝐵, 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑂𝐴 ≈ 𝑣 𝐵 𝑣𝐴 ∆𝜃 Velocity at 𝐵, parallel to 𝑂𝐴 = 𝑣𝑠𝑖𝑛∆𝜃 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑂𝐴 = 𝑟 𝑣∆𝜃 − 0 = 𝒗∆𝜽 ∆𝜃 If A is close to B, so that ∆𝜃 is small, 𝑂 𝑟 𝐴 𝑠𝑖𝑛∆𝜃 ≈ ∆𝜃 and 𝑐𝑜𝑠∆𝜃 ≈ 1 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑂𝐴, ⇒ 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝐵, 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑂𝐴 = 𝑣∆𝜃 =𝑣−𝑣 =𝟎 2  𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎 = 𝑡𝑖𝑚𝑒 𝑣∆𝜃 ∆𝜃 𝑎 = 𝑡 but 𝑡 =𝜔 𝒂 = 𝒗𝝎 𝑣 From 𝑣 = 𝜔𝑟 ⇒ 𝜔 = 𝑟 𝒗𝟐 ∴𝒂= 𝒓 Or 𝒂 = 𝝎𝟐 𝒓  CENTRIPETAL AND CENTRIFUGAL FORCES A body moving with a constant speed in a circular path experiences an outward force called centrifugal force. This force puts the body in a tendency of flying off along a tangent to the circular path. Centrifugal forces always act away from the center and are perpendicular to the direction of motion. In order for the body to continue moving in a circle without falling off, there must be an equal and opposite force to the centrifugal force. This force which counter balances the centrifugal force is called the centripetal force and always acts towards the center of the motion. Definition. Centripetal force is the force which keeps a body moving in a circular path and is directed towards the centre of the circular path. Consider a body of mass 𝒎, moving with speed 𝒗, in a circular path of radius 𝒓, the centripetal force is given by, 𝒗𝟐 𝑭 = 𝒎𝒂, where 𝒂 = 𝒓 is the centripetal acceleration 𝑚𝑣 2 ∴𝑭= 𝑜𝑟 𝐹 = 𝑚𝜔2 𝑟 𝑜𝑟 𝐹 = 𝑚𝜔𝑣 𝑟 Examples 1. A body of mass 0.2Kg moves with a constant angular speed of 5rads-1 around a circle of radius 10cm. Find the magnitude of the force that must be acting on the body towards the centre of the circle Solution + 𝐹 = 𝑚𝜔2 𝑟: 𝑚 = 0.2𝐾𝑔, 𝜔 = 5𝑟𝑎𝑑𝑠 −1 , 𝑟 = 10𝑐𝑚 = 0.1𝑚 𝑭 = 𝟎. 𝟐 × 𝟓𝟐 × 𝟎. 𝟏 𝑭 = 𝟎. 𝟓𝑵 2. A body moves round a circular track of radius 40m at 3 revolutions per second. Calculate i) The angular speed ii) The period iii) Speed of the body iv) Find also the angular speed of the body if it moves with uniform speed of 3ms-1 in a circle of radius 50cm. Solution. i) Angular speed, 𝜔 = 2𝜋𝑓; 𝑓 = 3 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 𝜔 = 2𝜋 × 3 𝝎 = 𝟔𝝅 𝒓𝒂𝒅𝒔−𝟏 1 ii) 𝑃𝑒𝑟𝑖𝑜𝑑, 𝑇 = 𝑓 1 𝑇=3 𝑻 = 𝟎. 𝟑𝟑𝒔 iii) 𝑆𝑝𝑒𝑒𝑑, 𝑣 = 𝜔𝑟 ; 𝜔 = 6𝜋 𝑟𝑎𝑑𝑠 −1 , 𝑟 = 40𝑐𝑚 = 0.4𝑚 𝑣 = 6𝜋 × 0.4 𝑣 = 2.4𝜋 𝑚𝑠 −1 3  iv) 𝑣 = 3𝑚𝑠 −1 , 𝑟 = 50𝑐𝑚 = 0.5𝑚, 𝜔 =? 𝑣 𝜔=𝑟 3 𝜔 = 0.5 𝝎 = 𝟔 𝒓𝒂𝒅𝒔−𝟏 Question 1. An insect trapped in a circular grove of radius 12cm moves along the grove steadily and completes 7 revolutions in 100 seconds. Find the angular speed and linear speed of motion. 2. A body of mass 0.1Kg is being revolved in a circular path of diameter 0.5m on a frictionless horizontal plane by means of a string. It performs 10 revolutions in 31.4 seconds. Calculate the centripetal force acting on the body. Examples of circular motion 1. An object whirled round at a constant speed in a horizontal circle by means of a string. The centripetal force in this case is equal to the tension in the string 𝑚 𝑇 𝑟 𝑇 = 𝑚𝜔2 𝑟 Tension,𝑇 in the string provides the centripetal force in this case. 2. An object whirled round at constant speed in a vertical circle by means of a string. In this case, tension in the string varies from position to position. 𝐴 If 𝑚 is the mass of the object, 𝑣 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑, and 𝑟 the length of the string 𝑚𝑔 (radius) then, 𝑇𝐴 𝑇𝐵 At position 𝐴 ( the top of the motion),𝑇𝐴 + 𝑚𝑔 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝐵 𝑚𝑣 2 𝑇𝐶 ⇒ 𝑇𝐴 + 𝑚𝑔 = 𝑟 𝑚𝑔 𝑚𝑣 2 𝐶 𝑇𝐴 = − 𝑚𝑔 …………..1 𝑟 𝑚𝑔 At position B, 𝑇𝐵 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑚𝑣 2 𝑇𝐵 = … … … … … … … … …..2 𝑟 Putting equation 2 into 1 𝑇𝐵 = 𝑇𝐴 + 𝑚𝑔 ⇒ 𝑇𝐴 ≤ 𝑇𝐵 At position C, 𝑇𝐶 − 𝑚𝑔 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑚𝑣 2 𝑇𝐶 − 𝑚𝑔 = 𝑟 ………………………..3 Putting equation 2 into 3, 4 𝑇𝐶 − 𝑚𝑔 = 𝑇𝐵 ⇒ 𝑇𝐵 ≤ 𝑇𝐶 ∴ 𝑻𝑨 ≤ 𝑻𝑩 ≤ 𝑻𝑪 It is therefore noticed that tension is maximum at the bottom of the motion (ie at C) and minimum at the top of the motion (i.e at A). Note; If a bucket containing some water is whirled in a vertical plane, it is found that at a certain speed the water does not pour out even at the top of the motion. This is possible when the weight, 𝑚𝑔 of water is less that the centripetal 𝑚𝑣 2 𝑚𝑣 2 force, 𝑟. However if 𝑚𝑔 is greater than 𝑟 , the water pours out and this occurs when the whirling is done at a low speed. Example 1. A stone of mass 0.5Kg is whirled round on the end of a 0.8m long string in a vertical circle. If the speed of the stone is 4ms-1. i) Determine the minimum tension in the string and state where it acts ii) Calculate the maximum tension Solution 𝑚𝑣 2 i) Minimum tension is at position A, 𝑇𝐴 + 𝑚𝑔 = 𝑟 ,𝑚 = 0.5𝐾𝑔, 𝑟 = 0.8𝑚, 𝐴 𝑣 = 4𝑚𝑠 −1 𝑚𝑔 𝑚𝑣 2 𝑇𝐴 𝑇𝐴 = − 𝑚𝑔 𝑟 0.5×42 𝑇𝐵 𝑇𝐴 = − 0.5 × 9.81 0.8 𝐵 𝑇𝐴 = 5.095𝑁 𝑚𝑔 ∴ 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒊𝒔 𝟓. 𝟎𝟗𝟓𝑵 𝑚𝑣 2 ii) Maximum tension is at B, 𝑇𝐵 − 𝑚𝑔 = 𝑟 𝑚𝑣 2 𝑇𝐵 = 𝑟 + 𝑚𝑔 0.5×42 𝑇𝐵 = 0.8 + 0.5 × 9.81 𝑻𝑩 = 𝟏𝟒. 𝟗𝟎𝟓𝑵 ∴ 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒊𝒔 𝟏𝟒. 𝟗𝟎𝟓𝑵 2. A particle of mass 0.5Kg attached to a string of length 0.5m which will break if the extension in it exceeds 20N. The particle is whirled in a vertical circle. The axis of rotation being at a height of 1.0m above the ground. The angular velocity is gradually increased until the string breaks. In what position is it most likely to break and at what angular speed? Determine the position at which the particle will hit the ground. 5  Solution It is most likely to break at A. 𝑚𝑣 2 𝑇𝐴 − 𝑚𝑔 = 0.5𝑚 𝑂 𝑟 2 𝑚𝑣𝑚𝑎𝑥 For maximum tension, 𝑇𝑚𝑎𝑥 = 𝑟 + 𝑚𝑔 0.5𝑚 𝑇𝐴 2 0.5𝑣𝑚𝑎𝑥 1.0𝑚 𝑣 20 = 0.5 + 0.5 × 9.81 𝐴 2 𝑣𝑚𝑎𝑥 = 20 − 4.905) 0.5𝑚 𝑚𝑔 𝑣𝑚𝑎𝑥 = 15.095 𝑣𝑚𝑎𝑥 = 3.89𝑚𝑠 −1 𝑥 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑, 𝜔 = 𝑣𝑟 𝜔 = 3.89 × 0.5 𝝎 = 𝟏. 𝟗𝟒 𝒓𝒂𝒅𝒔−𝟏 Position on the ground 1 𝑠𝑦 = 𝑢𝑦 𝑡 + 2 𝑎𝑦 𝑡 2 :𝑠 = 0.5𝑚, 𝑢𝑦 = 0, 𝑎𝑦 = 9.81𝑚𝑠 −2 1 0.5 = 0 + × 9.81𝑡 2 2 𝑡 = 0.319𝑠 1 𝑠𝑥 = 𝑢𝑥 𝑡 + 𝑎𝑥 𝑡 2 𝑠𝑥 = 𝑥, 𝑢𝑥 = 3.89𝑚𝑠 −1 , 𝑎𝑥 = 0, 𝑡 = 0.319𝑠 2 𝑥 = 3.89 × 0.319 𝒙 = 𝟏. 𝟐𝟒𝟐𝒎 Questions 1. An object of mass 3kg is whirled in a vertical circle of radius 2m with a constant sped of 12ms-1, calculate the maximum and minimum tension in the string 2. Explain why a bucket full of water can be swung round a vertical circle without spilling 3. A stone of mass 800g is attached to string of length 60cm which has a breaking tension of 20N. The string is whirled in a vertical circle the axis of rotation at a height of 100cm from the ground. i) What is the angular velocity where the string is most likely to break? ii) How long will it take before the stone hits the ground? 3. The conical pendulum A conical pendulum is formed by whirling a body round a horizontal circle at a constant speed by means of a string fixed at a point above the centre of the circle as shown below 𝑇 − 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑛𝑔 𝜃 𝑙 − 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑛𝑔 𝑇 𝑟 − 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑐𝑖𝑟𝑐𝑙𝑒 𝑑𝑒𝑠𝑐𝑟𝑖𝑏𝑒𝑑 𝑟 𝑚 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 𝑚𝑔 6 Resolving horizontally; 𝑚𝑣 2 𝑇𝑠𝑖𝑛𝜃 = ……………………..1 𝑟 Resolving vertically 𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔 ……………………..2 Dividing equations 1 by 2 𝑚𝑣2 𝑇𝑠𝑖𝑛𝜃 ⇒ = 𝑟 𝑇𝑐𝑜𝑠𝜃 𝑚𝑔 𝒗𝟐 ∴ 𝒕𝒂𝒏𝜽 = 𝒓𝒈 Examples 1. A body of mass 1.5Kg is attached to a string, 1m long and revolves round a horizontal circle of radius 0.4m, thus making a conical pendulum. Find the tension in the string and the period of the pendulum. Solution 𝑇 =?, 𝑙 = 1𝑚, 𝑟 = 0.4𝑚, 𝒗𝟐 𝑚 = 1.5𝐾𝑔,𝜃 =? From, 𝒕𝒂𝒏𝜽 = 𝜃 𝒓𝒈 𝑙 𝑟 𝑇 𝑠𝑖𝑛𝜃 = 𝑙 ⇒ 𝑣 = 0.4 × 9.81 × 𝑡𝑎𝑛23.6 𝑠𝑖𝑛𝜃 = = 0.4 0.4 𝑣 = 1.31𝑚𝑠 −1 1 2𝜋 𝑣 𝜃 = 23.6 0 But 𝜔 = = 𝑟 𝑇 𝑟 Resolving vertically 2×3.14×0.4 ⇒ 𝑇 = 1.31 𝑚𝑔 𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔 ∴ 𝑻 = 𝟏. 𝟗𝒔 𝑇𝑐𝑜𝑠23.6 = 1.5 × 9.81 𝑻 = 𝟏𝟔. 𝟎𝟔𝑵 2. A string which breaks under a tension of 12N is used to make a conical pendulum. If the string is 1m long and the bob is of mass 1Kg, find the maximum speed and frequency of the bob. 𝒗𝟐 𝑇 = 12𝑁, 𝑙 = 1𝑚, 𝑟 =? ,𝑚 = 1𝐾𝑔, 𝑣 =? From 𝒕𝒂𝒏𝜽 = 𝒓𝒈 𝜃 Resolving vertically, 𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔 𝑣 = 0.58 × 9.81𝑡𝑎𝑛35.2 𝑇 12𝑐𝑜𝑠𝜃 = 1 × 9.81 𝒗 = 𝟏. 𝟗𝟗𝟕𝒎𝒔−𝟏 𝜃 = 35.20 𝑣 𝑟 𝜔 = 2𝜋𝑓 = 𝑠𝑖𝑛35.2 = 1 𝑟 𝑟 1.997 𝑟 = 0.58 𝑚 𝑓 = 0.58×2×3.14 𝑚𝑔 𝒇 = 𝟎. 𝟓𝟓𝑯𝒛 Exercise 1. A 30Kg body is whirled in a horizontal circle as conical pendulum by means of an inelastic string that has a breaking strength of 392N. When the speed of the body is 8.0ms-1 the string breaks. Calculate i) The angle the string makes with the vertical at that instant ii) The length of the string 2. A stone of mass 0.3Kg tied to the end of a string in a horizontal plane is whirled round in a circle of radius 100cm with a frequency of 40 revolutions per minute. What is the tension in the string? Determine the maximum speed with which the stone can be whirled round if the string can withstand a maximum tension of 200N 3. A particle of mass 0.2Kg is attached to one end of a light inextensible string of length 50cm. the particle moves in a horizontal circle with angular velocity of 5.0 rad s-1 with the string inclined at an angle, 𝜃 to the vertical. Find the value of 𝜃 7 4. The period of oscillation of a conical pendulum is 2.0s. If the string makes an angle of 60° to the vertical at the point of suspension, calculate the: (i) Vertical height of the point of suspension above the circle.(h = 0.994m) (ii) Length of the string, (l = 1.99m) (iii) Velocity of the mass attached to the string.(v = 5.41ms-1) 5. A stone of mass 0.5kg is tied to one end of the string 1m long. The point of suspension of the string is 2m above the ground. The stone is whirled in the horizontal circle with increasing angular velocity. The string will break when the tension in it is 12.5N and the angle 𝜃 is to the maximum (𝜃𝑚𝑎𝑥)as shown in the figure below; 𝐴 i) Calculate the angle 𝜃𝑚𝑎𝑥 𝜃𝑚𝑎𝑥 ii) Calculate the angular velocity of the stone when the string breaks iii) How far from the point 𝐺 on the ground will the stone 2𝑚 hit the ground? 𝑂 iv) What will be the speed of the stone when it hits the ground? 𝐺 𝐺𝑟𝑜𝑢𝑛𝑑. MOTION OF A CAR ROUND A FLAT CIRCULAR TRACK [NEGOTIATING A BEND] a) Toppling/ overturning of the car Consider a vehicle of/ mass m moving with a speed v in a circle of radius r; let h be the height of the centre of gravity above the truck and 2a the distance between the tyres. 𝑅1 𝑅2 𝐹1 and 𝐹2 are frictional forces on the tyre and 𝑅1 and 𝑅2 are normal reactions at the point 𝑇𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝐺 of contacts of the wheels with the tyres. ℎ 𝑎 𝑎 For circular motion, the frictional forces 𝑂 𝐹1 𝐹2 𝐹1 and 𝐹2 provide the centripetal force. 𝑚𝑔 𝑚𝑣 2 𝐹1 + 𝐹2 = … … … … ….1 𝑟 Since the car does not move off the road, at Putting 𝑅1 in equation 4, equilibrium, 𝑚𝑣 2 𝑅2 − 𝑚𝑔 − 𝑅2 ) = ℎ 𝑅1 + 𝑅2 = 𝑚𝑔………………………………..2 𝑎𝑟 𝑚𝑣 2 Taking moments about G, 2𝑅2 = 𝑎𝑟 ℎ + 𝑚𝑔 𝑅1 × 𝑎 + 𝐹1 × ℎ + 𝐹2 × ℎ = 𝑅2 × 𝑎 𝟐 𝒎 𝒗 𝒉 𝑹𝟐 = +𝒈 𝑅2 − 𝑅1 )𝑎 = 𝐹1 + 𝐹2 )ℎ … … … …..3 𝟐 𝒂𝒓 Combining equations 1 and 3 Substituting for 𝑅2 from equation 2 into 4 𝑚𝑣 2 𝒎 𝒗𝟐 𝒉 𝑅2 − 𝑅1 )𝑎 = ℎ…………………4 𝑹𝟏 = 𝟐 𝒈− 𝒂𝒓 𝑟 From equation 2, 𝑅1 = 𝑚𝑔 − 𝑅2 … … 5 𝑹𝟐 Can never be zero, if 𝑹𝟏 = 𝟎, then the car is just about to overturn/topple/upset. 8  Note For no overturning, 𝑅1 ≥ 0 𝑅1 is the reaction of the inner tyre ⇒ 𝒎 𝒈− 𝒗𝟐 𝒉 ≥0 𝟐 𝒂𝒓 When 𝑅1 >0: The wheels in the inner side of the curve 𝒗𝟐 𝒉 𝒈≥ are in contact with the ground. 𝒂𝒓 𝒂𝒓𝒈 𝒗≤ √ 𝒉 When 𝑅1 = 0: The wheels in the inner side of the 𝑎𝑟𝑔 ∴ 𝑇ℎ𝑒 𝑐𝑎𝑟 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑜𝑣𝑒𝑟𝑡𝑢𝑟𝑛 𝑖𝑓 𝑣 ≤ √ curve are at the point of losing contact with the ground ℎ 𝒂𝒓𝒈 When 𝑅1 < 0: The inner wheels have lost contact For no overturning, 𝒗𝒎𝒂𝒙 = √ 𝒉 with the ground and the vehicle has over turned. Conditions for no overturning/toppling  The centre of gravity should be low i.e. height, ℎ should be small  The radius of the bend/ circular curve should be large  The distance between the two tyres should be large 𝒂𝒓𝒈  The vehicle should be moving at low speed, 𝒗 ≤ √ 𝒉 b) A skidding / sliding/slipping car If the coefficient between the tyres and the road surface is 𝜇 ⇒ 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝐹 = 𝜇𝑅 𝐹1 + 𝐹2 = 𝜇𝑅1 + 𝜇𝑅2 𝐹1 + 𝐹2 = 𝜇 𝑅1 + 𝑅2 ) ⇒ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 𝑖𝑛 (a) above becomes 𝑚𝑣 2 𝜇 𝑅1 + 𝑅2 ) = ……………….* 𝑟 𝐴𝑙𝑠𝑜 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2, 𝑹𝟏 + 𝑹𝟐 = 𝒎𝒈 ……………………** 𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ∗∗) 𝑖𝑛𝑡𝑜 ∗), 𝑚𝑣 2 𝜇𝑚𝑔 = 𝑟 ∴ 𝒗 = √𝝁𝒓𝒈 If the car is driven at maximum velocity, 𝒗𝒎𝒂𝒙 = √𝝁𝒓𝒈, then it is at the point of sliding/skidding/slipping Therefore if 𝒗 > √𝝁𝒓𝒈 , the car slides. Conditions for no skidding/sliding/slipping For a car to go round a bend successfully without skidding then:  The speed of the car should be low i.e. 𝒗 < √𝝁𝒓𝒈  The radius,𝑟 of the bend should be made big  Coefficient of friction,𝜇 should be increased  Centre of gravity should be low i.e. height ℎ should be small Banking of a track This is the building of a track round a corner with the outer edge raised above the inner one. This is done in order to increase the maximum safe speed for no skidding. When a road is banked, some extra centripetal force is provided by the horizontal component of the normal reaction 9 When determining the angle of banking during the construction of the road, friction is ignored. The angle through which the outer edge of the roads are raised is called the angle of banking Motion of a car round a banked track a) With no tendency to skid Consider a car of mass 𝑚 negotiating a banked track at a speed 𝑣 and radius of the bend is 𝑟. 𝑅2 𝑅1 To centre of circular path 𝑚𝑔 𝜃 For no tendency to slide/skid/slip at the wheels of the car, there are no frictional forces. The centripetal force is provided by the horizontal components of normal reaction, 𝑅1 and 𝑅2 𝑚𝑣 2 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 = 𝑟 ………………………………..1 At equilibrium vertically, 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 = 𝑚𝑔 ………………………………..2 Equation 1 divide by equation 2 𝑚𝑣2 𝑅1 +𝑅2 )𝑠𝑖𝑛𝜃 𝑅1 +𝑅2 )𝑐𝑜𝑠𝜃 = 𝑟 𝑚𝑔 𝒗𝟐 𝒕𝒂𝒏𝜽 = 𝒓𝒈 𝜃 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑏𝑎𝑛𝑘𝑖𝑛𝑔 b) With a tendency to slip/skid/slide Case 1. When the speed is maximum When the car is moving with speed 𝑣 such that the frictional forces on the inner and outer wheels are 𝐹1 and 𝐹2 respectively each acting towards the centre of the circular path (track). 𝑅2 𝑅1 To centre of circular path 𝐹2 𝐹1 𝑚𝑔 𝜃 Resolving forces horizontally, 𝑚𝑣 2 𝐹1 + 𝐹2 )𝑐𝑜𝑠𝜃 + 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 = 𝑟 but 𝐹 = 𝜇𝑅,⇒ 𝐹1 + 𝐹2 = 𝜇 𝑅1 + 𝑅2 ) 𝑚𝑣 2 ⇒ 𝜇 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 + 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 = …………………..1 𝑟 Resolving vertically 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 − 𝐹1 + 𝐹2 )𝑠𝑖𝑛𝜃 = 𝑚𝑔 ⇒ 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 − 𝜇 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 = 𝑚𝑔 …………………...2 Equation 1 divide 2 10  𝑚𝑣2 𝜇 𝑅1 +𝑅2 )𝑐𝑜𝑠𝜃+ 𝑅1 +𝑅2 )𝑠𝑖𝑛𝜃 𝑅1 +𝑅2 )𝑐𝑜𝑠𝜃−𝜇 𝑅1 +𝑅2 )𝑠𝑖𝑛𝜃 = 𝑟 𝑚𝑔 Divide each term by 𝑐𝑜𝑠𝜃 𝜇+𝑡𝑎𝑛𝜃 𝑣2 1−𝜇𝑡𝑎𝑛𝜃 = 𝑟𝑔 𝒓𝒈 𝝁+𝒕𝒂𝒏𝜽) ∴ 𝒗𝒎𝒂𝒙 = √ 𝟏−𝝁𝒕𝒂𝒏𝜽 This is the maximum speed a car can move round a banked road of radius 𝑟 under the influence of friction. Case 2. When the speed is minimum At minimum speed, the car has a tendency to slide towards the centre of the circular path, therefore the frictional force must act away from the centre as shown below. 𝑅2 𝑅1 To centre of circular path 𝐹2 𝐹1 𝑚𝑔 𝜃 Resolving forces horizontally, 𝑚𝑣 2 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 − 𝐹1 + 𝐹2 )𝑐𝑜𝑠𝜃 = 𝑟 but 𝐹 = 𝜇𝑅,⇒ 𝐹1 + 𝐹2 = 𝜇 𝑅1 + 𝑅2 ) 𝑚𝑣 2 ⇒ 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 − 𝜇 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 = 𝑟 𝑚𝑣 2 𝑅1 + 𝑅2 ) 𝑠𝑖𝑛𝜃 − 𝜇𝑐𝑜𝑠𝜃) = 𝑟 …………………..1 Resolving vertically 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 + 𝐹1 + 𝐹2 )𝑠𝑖𝑛𝜃 = 𝑚𝑔 ⇒ 𝑅1 + 𝑅2 )𝑐𝑜𝑠𝜃 + 𝜇 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 = 𝑚𝑔 𝑅1 + 𝑅2 ) 𝑐𝑜𝑠𝜃 + 𝜇𝑠𝑖𝑛𝜃) = 𝑚𝑔 …………………...2 Equation 1 divide by equation 2 𝑚𝑣2 𝑅1 +𝑅2 ) 𝑠𝑖𝑛𝜃−𝜇𝑐𝑜𝑠𝜃) = 𝑟 𝑅1 +𝑅2 ) 𝑐𝑜𝑠𝜃+𝜇𝑠𝑖𝑛𝜃) 𝑚𝑔 𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑏𝑦 𝑐𝑜𝑠𝜃 𝑡𝑎𝑛𝜃−𝜇 𝑣2 1+𝜇𝑡𝑎𝑛𝜃 = 𝑟𝑔 𝒓𝒈 𝒕𝒂𝒏𝜽−𝝁) ∴ 𝒗𝒎𝒊𝒏 = √ 𝟏+𝝁𝒕𝒂𝒏𝜽 Examples. 1. A bend of 200m radius on a level road is banked at the correct angle for a speed of 15ms-1. If a vehicle rounds the bend at 30ms-1, what is the minimum co-efficient of kinetic friction between the tyres and the road so that the vehicle will not skid? Solution 𝑣2 Angle of banking, 𝜃 = 𝑡𝑎𝑛−1 𝑟𝑔 152 𝜃 = 𝑡𝑎𝑛−1 200×9.81 𝜽 = 𝟔. 𝟓𝟎 11  𝒓𝒈 𝝁+𝒕𝒂𝒏𝜽) From 𝒗𝒎𝒂𝒙 = √ 𝟏−𝝁𝒕𝒂𝒏𝜽 200×9.81 𝜇+𝑡𝑎𝑛6.5) 30 = √ 1−𝜇𝑡𝑎𝑛6.5 302 1 − 𝜇𝑡𝑎𝑛6.5) = 1962 𝜇 + 𝑡𝑎𝑛6.5) 𝜇 1962 + 900𝑡𝑎𝑛6.5) = 900 − 1962𝑡𝑎𝑛6.5 900−1962𝑡𝑎𝑛6.5 𝜇= 1962+900𝑡𝑎𝑛6.5) 𝝁 = 𝟎. 𝟑𝟐𝟕 2. A car travels round a bend in road which is a circular arc of radius 62.5m. 5 The road is banked at angle 𝑡𝑎𝑛−1 to the horizontal the coefficient of friction between the tyres of the car 12 and the road surface is 0.4. Find (i) The greatest speed at which the car can be driven round the bend without slipping. (ii) The least speed at which the car can be driven round the bend without slipping. Solution 𝒓𝒈 𝝁+𝒕𝒂𝒏𝜽) 5 i) 𝒗𝒎𝒂𝒙 = √ 𝟏−𝝁𝒕𝒂𝒏𝜽 , 𝐿𝑒𝑡 𝜃 𝑏𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑏𝑎𝑛𝑘𝑖𝑛𝑔 ⟹ 𝑡𝑎𝑛𝜃 = 12 𝟓 𝟔𝟐.𝟓×𝟗.𝟖𝟏 𝟎.𝟒+ 𝑣𝑚𝑎𝑥 = √ 𝟓 𝟏𝟐 𝟏−𝟎.𝟒× 𝟏𝟐 Greatest speed, 𝒗𝒎𝒂𝒙 = 𝟐𝟒. 𝟓𝟏𝒎𝒔−𝟏 𝒓𝒈 𝒕𝒂𝒏𝜽−𝝁) ii) 𝒗𝒎𝒊𝒏 = √ 𝟏+𝝁𝒕𝒂𝒏𝜽 𝟓 𝟔𝟐.𝟓×𝟗.𝟖𝟏 −𝟎.𝟒) 𝒗𝒎𝒊𝒏 = √ 𝟏𝟐 𝟓 𝟏+𝟎.𝟒× 𝟏𝟐 𝒗𝒎𝒊𝒏 = 𝟐. 𝟗𝟔𝒎𝒔−𝟏 3. A road banked at 100 goes round a bend of radius 70m. At what speed can a car travel round the bend without tending to side slip. Solution 𝑣2 𝑡𝑎𝑛𝜃 = 𝑟𝑔 𝑣 = 70 × 9.81𝑡𝑎𝑛10 𝒗 = 𝟏𝟏. 𝟎𝒎𝒔−𝟏 Exercise 1. A racing car of mass 2 tonnes is moving at a speed of 5ms-1 round a circular path. If the radius of the track is 100m. calculate; i) Angle of inclination of the track to the horizontal if the car does not tend to side slip ii) The reaction to the wheel if it’s assumed to be normal to the track 2. A car travels round a curved road banked at an angle of 22.6°. If the radius of curvature of the bend is 62.5m and the coefficient of friction between the tyres of the car and the road surface is 0.3. Calculate the maximum speed at which the car negotiates the bend without skidding. 12 3. On a level race track, a car just goes round a bend of radius 80m at a speed of 20ms-1 without skidding. At what angle must the track be banked so that a speed of 30ms-1 can just be reached without skidding, the coefficient of friction being the same in both cases? 4. A car of mass 1000kg moves round a banked track at a constant speed of 108km/h. Assuming the total reaction at the wheels is normal to the track, and the radius of curvature of the track is 100m, calculate the; (i) Angle of inclination of the track to the horizontal. (ii) Reaction at the wheels. Motion of a bicycle rider round a circular track. Consider a rider on a bicycle as shown below 𝑅 The frictional force, 𝐹 has a turning effect on the rider about the centre of gravity, G. 𝑇𝑜 𝑐𝑒𝑛𝑡𝑟𝑒 𝑮 In order to balance this effect the rider has to lean ℎ 𝜃 inwards at an angle 𝜃 to the vertical, so that the turning effect or moment due to normal reaction, 𝑅 about 𝐹 𝑎 centre of gravity, G counter balances that due to 𝑚𝑔 frictional force, 𝐹 to avoid toppling. a) For no skidding; 𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦; 𝑅 = 𝑚𝑔 ………………………………….1 The frictional force provides the centripetal force 𝑚𝑣 2 𝐹= 𝑟 𝐵𝑢𝑡 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝐹 = 𝜇𝑅 𝑚𝑣 2 ⇒ 𝜇𝑅 = 𝑟 ……………………………2 𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 𝑖𝑛𝑡𝑜2 𝑚𝑣 2 𝜇𝑚𝑔 = 𝑟 𝒗 = √𝝁𝒓𝒈 Where 𝜇 − the coefficient of sliding friction between the tyres and the road is surface, 𝑟 is the radius of the circular track and 𝑣 is the maximum speed the cyclist negotiates the circular bend without skidding/. 𝒗𝒎𝒂𝒙 = √𝝁𝒓𝒈 b) For no toppling. A force, 𝐹 has a moment 𝐹ℎ about 𝐺 tends to turn the rider outwards. When the rider leans as shown it is counter balanced by the moment, 𝑅𝑎 of 𝑅 about 𝐺, Taking moments about G 𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦; 𝐹ℎ = 𝑅𝑎 𝑅 = 𝑚𝑔 ……………3 𝑎 𝐹 Combining equations 1, 2 and 3, ⟹ = ……1 ℎ 𝑅 𝑚𝑣2 The frictional force provides the 𝑎 𝑎 ⇒ ℎ = 𝑟 𝑚𝑔 𝑏𝑢𝑡 ℎ = 𝑡𝑎𝑛𝜃 centripetal force 𝒗𝟐 𝑚𝑣 2 𝒕𝒂𝒏𝜽 = 𝒓𝒈 𝐹= 𝑟 ………….2 𝒗 = √𝒓𝒈𝒕𝒂𝒏𝜽 ; 𝑣 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑡ℎ𝑒 𝑐𝑦𝑐𝑙𝑖𝑠𝑡 𝑛𝑒𝑔𝑜𝑡𝑖𝑎𝑡𝑒𝑠 𝑎 𝑐𝑜𝑟𝑛𝑒𝑟 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑡𝑜𝑝𝑝𝑙𝑖𝑛𝑔 13  For no toppling 𝑣 2 ≤ 𝑟𝑔𝑡𝑎𝑛𝜃. Thus condition for skidding N.B 𝒗𝟐 An areoplane will move with a speed 𝑣 𝑚𝑠 −1 in a horizontal circular path of radius 𝑟 such that 𝑡𝑎𝑛𝜃 = if its 𝒓𝒈 wing are banked at an angle 𝜃 to the horizontal Question 1. Why it is necessary for a bicycle rider moving round a circular path to lean towards a center of the path 𝑅 𝑇𝑜 𝑐𝑒𝑛𝑡𝑟𝑒 𝑮 𝜃 ℎ 𝐹 𝑎 𝑚𝑔 When a person on a bicycle is riding round a circular path, the frictional force at the ground provides the centripetal force. This force provides a moment, 𝐹𝒉 about his centre of gravity. The rider would therefore have a tendency to fall off in a direction away from the centre of the path if this moment is not counter balanced. However when the rider leans towards the centre of the path, his normal reaction bears a moment, 𝑹𝒂 about his centre of gravity, which counter balances the moment due to friction. 2. A bicycle rider is moving round a bend of radius 1.5m at a speed of 6Kmh-1. Find the coefficient of friction between the bicycle tyres and road surface for the bicycle not to slide-slip. Solution 𝑣 = 6𝑘𝑚ℎ−1 = 1.6667𝑚𝑠 −1 , 𝑟 = 1.5𝑚 𝐹𝑟𝑜𝑚 𝑣 = √𝜇𝑟𝑔 1.66672 𝜇 = 1.5×9.81 𝜇 = 0.2 3. Explain why the maximum speed of a car on a banked road is higher than that on an unbanked road. 𝑅1 𝑅2 𝑇𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝐺 𝑂 𝐹1 𝐹2 𝑚𝑔 When the track is unbanked, it’s only the frictional force, 𝐹 acting on the wheels that provides the centripetal 𝑚𝑣12 force (i.e. 𝐹1 + 𝐹2 = 𝑟 ). Such that when limiting frictional force is exceeded, slipping occurs. 𝑅2 𝑅1 To centre of circular path 𝐹2 𝐹1 𝑚𝑔 𝜃 14  When the track is banked, the normal reaction has a component towards the centre of the circular track. Therefore the centripetal force is provided by the component of friction, together with this component of the 𝑚𝑣22 normal reaction ( 𝐹1 + 𝐹2 )𝑐𝑜𝑠𝜃 + 𝑅1 + 𝑅2 )𝑠𝑖𝑛𝜃 = 𝑟 ). 𝑚𝑣22 𝑚𝑣12 This centripetal force is larger than when the track is unbanked, i.e 𝑟 > 𝑟 ⇒ 𝑣2 > 𝑣1. , and so a racing car can travel faster on a banked track than on a flat track. 4. Explain why a mass attached to a string rotating at a constant speed in a horizontal circle will fly off at a tangent if the string breaks. The velocity of a body executing uniform circular motion is always tangent to the circular path. The centripetal force causing the mass along the horizontal circle is provided by the tension in the string. So when the string breaks there is no more centripetal force and so the mass will fly off at a tangent. 5. Explain why a force is necessary to maintain a body moving with constant speed in a circular path A body moving in a circular path at constant speed constantly changes direction and hence its velocity changes. According to Newton’s first law of motion, a change of velocity is evidence of a force. 15