Rotational Kinematics PDF

Rotational Kinematics SPH 3107: PHYSICS 1A Lecture 1 Circular Motion After studying this lesson, you should: appreciate that circular motion requires a force to sustain it; know and understand tha...

Rotational Kinematics SPH 3107: PHYSICS 1A Lecture 1 Circular Motion After studying this lesson, you should: appreciate that circular motion requires a force to sustain it; know and understand that, for motion in a circle with uniform angular velocity, the acceleration and the force causing it are directed towards the centre of the circle; use your knowledge to model applications of circular motion. 2  …. The heavenly bodies, the planets and the stars, moved in circles around the Earth. Once set upon their paths by the Gods, these bodies continued to move in circles without any further intervention. No force was required to sustain their heavenly orbits. However, from Newton’s point of view, circular motion requires a force to sustain it. Recall Newton’s First Law: A body remains in a state of rest or moves with uniform motion, unless acted upon by a force. A body describing a circle is certainly not at rest. Nor does it move with constant velocity, since, whether the speed is changing or constant, the direction of the motion is changing all the time. From Newton’s First Law, there must be a resultant force acting on the body and Newton’s Second Law equates this force to the mass times acceleration, so the body must be accelerating. 3  Remember back… There are 3 ways that an object can You already know some of this… be experiencing acceleration? If each blade in the wind farm animation is 30 meters long, estimate the speed (in ms-1) of the tip of one turbine blade. Speeding Up Slowing Down 𝑑 188.5 𝑚 𝑣= = = 65.0 𝑚𝑠 −1 𝑡 2.9 𝑠 Changing Direction 4  Angular displacement, radian, period… If you walked around this circle If a child on a merry-go-round Period (T): Time for once, what is your total distance? rotates 3.5 times, what is their complete revolution angular distance in radians? In circular motion, angular 3.5 2𝜋 = 7𝜋 = 22 𝑟𝑎𝑑 displacement is defined as the change in the angle of a If an ant on a record player spins for body with respect to its 𝐶 = 2𝜋𝑟 initial angular position. an angular displacement of 14 radians, how many revolutions has How many radians are it experienced? therein one full revolution? 14 𝟐𝝅 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 = 2.23 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑠 = 𝑟𝜃 2𝜋 5  Angular Velocity The rate of change of angular displacement is called the 𝑑𝜃 𝑣 angular velocity of the particle. ∴ = 𝑑𝑡 𝑟 𝑑𝜃 𝜔= 𝑑𝑡 𝑣 For one complete revolution, the ∴𝜔= 𝑟 angle swept by the radius vector is Let PQ = ds, be the 360𝑜 or 2π radians. arc length If T is the time taken for one 𝑣 = 𝑟𝜔 complete revolution, known as 𝑑𝑆 𝑑𝜃 = period, then the angular velocity of 𝑟 Relation between linear the particle is But 𝑑𝑠 = 𝑣𝑑𝑡 velocity and angular velocity 2𝜋 𝑣𝑑𝑡 𝜔= ∴ 𝑑𝜃 = 𝑇 𝑟 6  Angular Acceleration The rate of change of angular velocity with Example: A powerful motorcycle can time is called angular acceleration. accelerate from 0 to 30.0 m/s in 4.20 s. 𝑑𝜔 𝑑 𝑑𝜃 𝑑2 𝜃 What is the angular acceleration of its 𝛼= = = 2 0.320-m-radius wheels? 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑 𝑡 𝑑𝑣 30 𝑎𝑡 = = = 7.14 𝑚𝑠 −2 𝜔2 − 𝜔1 𝑑𝑡 4.2 𝛼= 𝑡 𝑎𝑡 7.14 𝛼= = = 22.3 𝑟𝑎𝑑/𝑠 2 The angular acceleration is measured in terms of 𝑟 0.32 𝑟𝑎𝑑 𝑠 −2 and its dimensional formula is 𝑇 −2. Exercise: A particle moves round a circle 𝑑𝑣 𝑑(𝑟𝜔) 𝑑𝜔 𝑎𝑡 = = =𝑟 in 10 seconds at a constant speed of 15 𝑑𝑡 𝑑𝑡 𝑑𝑡 ms-1. Calculate the angular speed of the particle and the radius of the circle. 𝑎𝑡 = 𝑟𝛼 7  Activities My washing machine has a spin speed of 1200 rpm, and a drum radius of 25 cm. Calculate the linear speed of clothes at the edge of the drum. A particle moves along a circular path of radius 4.0 m with an angular speed of 30 rads-1. (a) Calculate the linear speed of the particle. (b) Calculate the frequency in revolutions per second. (c) Calculate the time taken for one revolution. An airplane propeller is rotating with uniform speed of 1,800 rpm. The blades of the propeller are 6 m long. Determine the linear speed of a point (a) 2 m from the axis and (b) 6m from the axis. Expressing angular speed in radians per second 2𝜋 × 1800 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 𝑟𝑎𝑑 𝜔= = 188.4 60 𝑠 𝑠 𝑣 = 𝑟𝜔 = 2 × 188.4 = 376.8 𝑚/𝑠 8  Linear motion vs. Circular motion equations Example: A body moves on a circular path of radius 20 𝑐𝑚 at 𝑑𝑣 6 − 5 a speed that uniformly 𝑎 𝑡 = = = 0.5 𝑚𝑠 −2 𝑑𝑡 2 increases. If the speed changes 𝛼 = 𝑎𝑡 = 0.5 = 2.5 𝑟𝑎𝑑 𝑠 −1 from 5 𝑚𝑠 −1 to 6 𝑚𝑠 −1 in 2 𝑠, 𝑟 0.2 find the angular acceleration. Example: A particle is moving in a circle of radius 2 𝑚 at a speed given by 𝑣 = 4𝑡, where v is in 𝑚𝑠 −1 and t is in seconds. (a) Calculate the tangential acceleration at 𝑡 = 1 𝑠. Exercise: If angular displacement of (b) Find total acceleration at 𝑡 = 1 𝑠. a particle is given by 𝜃 = 𝑎 − 𝑏𝑡 + 𝑑𝑣 𝑑 𝑣 2 42 𝑐𝑡 2 , then find its angular velocity 𝑎𝑡 = = 4𝑡 = 4 𝑚𝑠 −1 𝑎𝑐 = = = 8 𝑚𝑠 −2 𝑑𝑡 𝑑𝑡 𝑟 2 and angular acceleration. 𝑎= 𝑎𝑡2 + 𝑎𝑐2 = 4 5 𝑚𝑠 −2 9  Examples… You are playing fetch with your dog. You throw a ball A pulley rotating in the counterclockwise direction with an initial angular speed of 36.0 rad/s and your is attached to a mass suspended from a string. The dog catches the ball 0.595 s later. When your dog mass causes the pulley’s angular velocity to catches the ball, its angular speed has decreased to decrease with a constant angular acceleration 34.2 rad/s due to air resistance. 𝛼 = -2.10 rad/s2. (a) What is the ball’s angular acceleration, assuming it (a) If the pulley’s initial angular velocity is 5.40 to be constant? (b) How many revolutions does the ball make before rad/s, how long does it take for the pulley to being caught? come to rest? (b) Through what angle does the pulley turn during this time? 10  Centripetal Acceleration The fact that the direction of motion of an object in a circular motion changes continuously with time suggests that the velocity vector of the object varies continuously with time. Hence the motion must have an acceleration, a. We call accelerations giving rise to circular motions centripetal accelerations. 𝑃𝑄 + 𝑄𝑆 = 𝑃𝑆 −𝑣1 + 𝑣2 = ∆𝑣 ∆𝑣 = 𝑣2 − 𝑣1 The triangle PQS and AOB are similar. Therefore, ∆𝑣 𝑣 𝑣2 = 𝑎= 𝐴𝐵 𝑟 𝑟 𝐴𝐵 = 𝑎𝑟𝑐 𝐴𝐵 = v∆𝑡 ∆𝑣 𝑣 ∆𝑣 𝑣 2 = ⇒ = 𝑣∆𝑡 𝑟 ∆𝑡 𝑟 11  …. Example: A particle is moving with a constant Exercise: A deep-sea fisherman hooks a big fish angular acceleration of 4 𝑟𝑎𝑑 𝑠 −1 on a that swims away from the boat pulling the circular path. Initially the particle was at rest. fishing line from his fishing reel. The whole At what time, the magnitudes of centripetal system is initially at rest and the fishing line acceleration and tangential acceleration are unwinds from the reel at a radius of equal. 4.50 𝑐𝑚 from its axis of rotation. The reel is Tangential acceleration 𝑎𝑡 = 𝑟𝛼 given an angular acceleration of 110𝑟𝑎𝑑 𝑠 −2 Eqn. of motion 𝑣 = 0 + 𝑟𝛼𝑡 for 2.00 𝑠. 𝑣2 𝑟 2 𝛼2 𝑡 2 (a) What is the final angular velocity of the reel? Centripetal acceleration 𝑎𝑐 = = (b) At what speed is fishing line leaving the reel 𝑟 𝑟 after 2.00 s elapses? On equating |𝑎𝑡 | = |𝑎𝑐 | (c) How many revolutions does the reel make? 𝑟2𝛼 2𝑡2 (d) How many meters of fishing line come off rα = the reel in this time? 𝑟 2 1 1 1 𝑡 = = ⇒ 𝑡= 𝑠 𝛼 4 2 12  Centripetal Force Centripetal force Example: Calculate the tension force required to keep a 1.5 This is the force which keeps the body kg mass in circular motion at the end of a 2.0 m piece of string with a frequency of 80 rpm. moving in a circular path and it is 𝑅𝑃𝑀 80 directed towards the center of the 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = = = 1.33 𝐻𝑧 60 60 circular path. 𝑚𝑣 2 𝑇 = 𝑚𝑎 = = 𝑚𝜔2 𝑟 𝑟 𝑇 = 𝑚(2𝜋𝑓)2 𝑟 𝑇 = 4𝜋 2 × 1.5 × 2.0 × 1.332 = 210 𝑁 Activity: A child twirls his yo-yo about his head rather than using it properly. The yo-yo has a mass of 0.200 kg and is attached to a string 0.800 m long. If the “yo yo” 𝑣2 𝐹𝑐 = 𝑚𝑎𝑐 = 𝑚 = 𝑚𝜔2 𝑟 makes a complete revolution each second, what tension 𝑟 must exist in the string? Centrifugal force 4𝜋 2 𝑟 𝑇=𝑚 2 This is the force acting on a body 𝑇 moving in a circular path. 13  Centripetal Force… Example: A plane comes out of a power dive, turning upward in a curve whose center of curvature is 1300 m above the plane. The plane's speed is 260 m/s. a) Calculate the upward force of the seat cushion on the 100 kg pilot of the plane. A 10.0 kg block rests on a frictionless surface and is attached to a vertical peg by a rope. What is the tension in the rope if the block is whirling in a b) Calculate the upward force on a 90.0 g sample of blood in horizontal circle of radius 2.00 m with a linear the pilot's head. speed of 20 m/s? What would be the tension in the rope at the top and the bottom of the swing if it were whirled in a vertical circle? 14  Friction Def: When a body slide or tends to slide on a surface on which it is resting, a resisting force Types of Friction opposing the motion is produced at the contact Static friction: Friction experienced by a surface. This restoring force is called friction body when it is at rest is called static or frictional force. friction. Dynamics friction: Friction experienced by a body when it is in motion is called dynamic friction. Sliding friction: Friction experienced by a body when it slides over another body is called sliding friction. Rolling Friction: Friction experienced Limiting friction: the maximum force that can by a body when it rolls over another be developed at the contact surface, when the body is called rolling friction. body is just on the point of moving is called limiting force of friction. 15  Friction Frictional force 𝐹𝑟 depends on the normal reaction 𝑅 = 𝑚𝑔 𝐹𝑟 ∝ 𝑅 𝐹𝑟 = 𝜇𝑅 where 𝜇 is the coefficient of friction, 𝜇 depends on the nature of two surfaces that are in relative motion. For a body resting on a rough surface, when 𝐹𝑠𝑙𝑖𝑑𝑖𝑛𝑔/𝑘𝑖𝑛𝑒𝑡𝑖𝑐 = 𝐹𝑘 = 𝜇𝑘 𝑅 external force (F) is applied on it (pushing where 𝜇𝑘 is coefficient of /pulling), then this force has to overcome kinetic/sliding friction. friction (𝐹𝑟 ) before the body moves. The 𝐹𝑠𝑡𝑎𝑡𝑖𝑐 = 𝐹𝑠 ≤ 𝜇𝑠 𝑅 body is moved by a net force (𝐹𝑛𝑒𝑡 ) i.e. where 𝜇𝑠 is coefficient of static 𝐹𝑛𝑒𝑡 = 𝐹 − 𝐹𝑟 friction. 16  Friction Laws of Dynamic Friction The friction force always acts in a direction, opposite that in which the body is moving. The ration of limiting force and normal reaction is constant, and it is known as coefficient of friction. For moderate speeds, the friction force remains constant. But it decreases slightly Laws of Static Friction with the increase of speed. The friction force always acts in a direction, opposite to that in which the body tends to Example: A block of wood of mass 20 kg requires a horizontal force of 50 N to pull it with a uniform move. velocity along a horizontal surface. Calculate the The magnitude of friction force is equal to the coefficient of friction between the block and the surface. external force. 𝑅 = 𝑚𝑔 = 20 × 10 = 200𝑁 The ratio of limiting force and normal 𝐹𝑟 = 𝜇𝑅 reaction is constant. 𝜇 = 50ൗ200 = 0.25 17  Friction The frictional force is the product of the normal reaction Example: A mass of 5kg is placed on a plane force and the coefficient of friction: inclined at an angle of 300 to be horizontal. 𝐹𝑟 = 𝜇𝑅 = 0.5 × 43.3 = 21.65 𝑁 Calculate the force required to pull the mass up The component of the gravitational force pulling the the plane at uniform velocity if 𝜇 = 0.5. object down the incline is: 𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 𝑚𝑔𝑠𝑖𝑛 30 = 50 × 0.5 = 25 𝑁 There are two forces opposing motion, 𝐹𝑟 and 𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 Since the object is moving at a uniform velocity, the net force is zero. The applied force must balance both the frictional force and the component of the gravitational force acting down the incline. 𝐹𝑛𝑒𝑡 = 𝐹 − 𝐹𝑟 + 𝑚𝑔𝑠𝑖𝑛𝜃 The normal reaction force is the component of the We have uniform motions; net force is zero object's weight acting perpendicular to the inclined plane. It is given by: 𝐹 = 𝐹𝑟 + 𝑚𝑔𝑠𝑖𝑛30 = 21.65 + 25 = 46.25 𝑁 𝑅 = 𝑚𝑔 𝑐𝑜𝑠30𝑜 = 50 𝑐𝑜𝑠30 = 43.3 𝑁 18  Friction Example: A pull of 50 N inclined at 30o to The normal force N is the force exerted by the the horizontal is necessary to move a wooden table that balances the downward gravitational block on horizontal table. If coefficient of force (the weight of the block) minus the friction is 0.20, find the weight of the upward vertical component of the applied force. wooden block. If W is the weight of the block, the normal force is: 𝑁 = 𝑊 − 𝐹𝑣𝑒𝑡𝑖𝑐𝑎𝑙 = 𝑊 − 25 The frictional force 𝐹𝑟​ is given by the product of the coefficient of friction 𝜇 and the normal The applied force 𝐹=50 N is at an angle of force 𝑁 30o to the horizontal. So, we need to resolve 𝐹𝑟 = 𝜇. 𝑁 = 0.20(𝑊 − 25) this force into horizontal and vertical 𝐹ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 = 𝐹𝑟 components: 43.3 = 0.20 𝑊 − 25 𝐹ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 = 𝐹𝑐𝑜𝑠30 = 43.3 𝑁 𝑊 = 241.5 𝑁 𝐹𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 = 𝐹𝑠𝑖𝑛30 = 25 𝑁 19 Activity A wooden block of weight 50N rests on a horizontal plane. Determine the force required which is acted at an angle of 150 to just (i) Pull it, and (ii) Push it. Take coefficient friction = 0.4 between the mating surfaces. Comment on the result. 20 Application of Circular Motion: Banking of a curved road 21  Banking of roads…. 1. Vehicle Along a Horizontal Circular Track 2. Vehicle Along a Banked Frictionless Road Exercise: An automobile corner curve of radius R at a speed v. In terms of R and v and any other required physical constants, what is the minimum coefficient of friction required for the turn? 22  Banking of roads…. Example: A curve has a radius of 50 meters 3. Vehicle Along a Banked Road with Friction with a banking angle of 37°. What will be the ideal or critical velocity for a car on this curve using Banking Road Formula ? v² = (rg tanθ) ∴ v² = 50 × 10 × tan 37° ∴ v² = 500 × 3/4 ∴ v² = 1500 / 4 ∴ v² = 375 ∴ v = 19.36 m/s Activity: Banked curves are often used on racetracks to enable cars to safely execute turns at high speeds. Assuming no friction to help the car stay in a circular curve, what is the maximum speed that a 2000 kg car can travel around a curve of radius 50.0 m if the angle at which the curve is banked is 25° above the horizontal? With no friction, what will happen to a car if it goes slower than this speed? What if it goes faster? Explain! 23  Other applications…. Example: A racing track of curvature 9.9 m is 4. Conical Pendulum banked at tan⁻¹(0.5). Coefficient of static friction between the track and the tyres of a vehicle is 0.2. Derivation of formula for Time period Determine the minimum speed of vehicle so that it does not slip inwards. 24 Rotational Kinetic Energy and Moment of Inertia 𝑛 When we look at rolling motion, it is a 1 combination of both translational motion and 𝐾𝐸𝑡 = ෍ 𝑚𝑖 𝑣𝑖2 2 rotational motion. 𝐼=1 So, think of an object of radius 𝑟 that rolls or rotates with an angular speed of 𝜔, and Since we already know that 𝑣 = 𝑟𝜔 𝑛 translates or moves with a linear speed of 𝑣. 1 𝐾𝐸𝑟 = ෍ 𝑚𝑖 𝜔2 𝑟𝑖2 That means we need to consider the kinetic 2 𝐼=1 energy in both the linear and rotational worlds and add them up to get the total kinetic energy 𝑛 in an object that rolls and moves at the same 1 𝐾𝐸𝑟 = ෍ 𝑚𝑖 𝑟𝑖2 𝜔2 time. 2 𝑖=1 Translational kinetic energy is given as: 𝑛 Moment of inertia: The 1 1 1 Moment of 𝐼 = ෍ 𝑚𝑖 𝑟𝑖2 measure of an 𝐾𝐸𝑡 = 𝑚1 𝑣1 + 𝑚2 𝑣2 + ⋯ + 𝑚𝑛 𝑣𝑛 2 2 2 Inertia object’s 2 2 2 𝑖=1 resistance to 𝑛 is the total number of particles the rigid 1 2 𝐾𝐸𝑟 = 𝐼𝜔 change angular body has been subdivided into. 2 motion. 25  … Example: A 1.20 kg disk with a radius of 10 cm The moment of inertia of a body depends on its rolls without slipping. If the linear speed of the mass, size, and shape, and also on a particular disk is 1.41 m/s, find: axis around which the body is rotated. (a) The translational kinetic energy of the disk Exercise: The moment of inertia of a 0.98 kg wheel (b) The rotational kinetic energy of the disk rotating about its center is 0.13 kg·m2. What is the (c) The total kinetic energy of the disk radius of the wheel, assuming the weight of the spokes 1 2 1 2 can be ignored? 𝐾𝐸𝑡 = 𝑚𝑣 = 1.2 0.1 = 1.193 𝐽 Exercise: A system of point particles is shown in the 2 2 following figure. Each particle has a mass of 0.3 kg and 1 2 1 1 2 𝑣 2 1 they all lie in the same plane. What is the moment of 𝐾𝐸𝑟 = 𝐼𝜔 = 𝑚𝑟 = 𝑚𝑣 2 inertia of the system about the given axis? 2 2 2 𝑟 4 1 𝐼 = ෍ 𝑚𝑖 𝑟𝑖2 = 𝑚 ෍ 𝑟𝑖2 𝐾𝐸𝑟 = 1.2 1.41 2 = 0.596 𝐽 4 𝐼 = 0.5 0.62 + 0.42 + 0.22 𝐾𝐸 = 𝐾𝐸𝑡 + 𝐾𝐸𝑟 = 1.193 + 0.596 = 1.789 𝐽 𝐼 = 0.168 𝑘𝑔𝑚2 26  Torque Forces cause accelerations What cause angular accelerations? A door is free to rotate about an axis through O. There are three factors that determine the effectiveness of the force in loosening the tight bolt: ▪ The magnitude of the force ▪ The position of the application of the force ▪ The angle at which the force is applied Definition: Torque, 𝜏, is the tendency of To make an object start rotating, a force is a force to rotate an object about some needed; the position and direction of the force axis. matter as well. 27  Torque Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force, with F perpendicular to r. Torque will have direction… If the turning tendency of the force is counterclockwise, the torque will be positive. If the turning tendency is clockwise, the torque will be negative. The SI units of torque are N.m. Torque is a vector quantity 𝜏Ԧ = 𝑟Ԧ × 𝐹Ԧ In magnitude, torque is given as 𝜏 = 𝑟𝐹 sin 𝜙 = 𝐹𝑙 28  Torque: Understanding 𝒔𝒊𝒏 𝜽 The component of the force 𝐹 cos 𝜙 has no tendency to produce a rotation. The component of the force 𝐹 sin 𝜙 causes it to rotate. The moment arm, 𝑙, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force, 𝑙 = 𝑟 sin 𝜙 𝜏 = 𝑟𝐹 sin 𝜙 = 𝐹𝑙 29  Net Torque The force 𝐹1 will tend to cause a counterclockwise rotation about O. The force 𝐹2 will tend to cause a clockwise rotation about O. ෍ 𝜏 = 𝜏1 + 𝜏2 + 𝜏3 = 𝐹1 𝑙1 + −𝐹2 𝑙2 + 0 𝐼𝑓 σ 𝜏 ≠ 0, starts rotating 𝐼𝑓 σ 𝜏 = 0, rotation rate does not change Quiz: Under what conditions can a rotating body be in equilibrium? Ans: The total torque is zero, the angular acceleration is zero, and angular velocity is zero. 30  Activities…. Exercise: What three factors affect the torque Exercise: When opening a door, you push on it created by a force relative to a specific pivot perpendicularly with a force of 55 𝑁 at a distance of point? (3 marks) 0.85 𝑚 from the hinges. a) What torque are you exerting relative to the Exercise: A wrecking ball is being used to hinges? knock down a building. One tall unsupported b) Does it matter if you push at the same height as concrete wall remains standing. If the wrecking the hinges? ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by Exercise: If a 50 kg boy sits 3 m left of the center of the see-saw, and a 40kg girl sits 5 m right of the falling straight down? Explain your answer. center of the see-saw, where should the see-saw be (3 marks) supported if the see-saw has a mass of 70kg and a Exercise: Mechanics sometimes put a length of length of 10m? pipe over the handle of a wrench when trying to remove a very right bolt. How does this help? (It is hazardous since it can break the bolt). Ans: increase torque on the bolt. 31  Activities…. Example: The two children shown in the Figure (a) 𝜏 = 𝑟𝐹 𝑠𝑖𝑛 𝜃 𝜏2 = −𝜏1 below are balanced on a seesaw of negligible 𝜃 = 90𝑜 𝑟2 𝑊2 = 𝑟1 𝑊1 mass. The first child has a mass of 26.0 𝑘𝑔 and 𝜏1 = 𝑟1 𝑊1 𝑟2 𝑚2 𝑔 = 𝑟1 𝑚1 𝑔 sits 1.60 𝑚 from the pivot. 𝜏2 = −𝑟2 𝑊2 𝑚1 a) If the second child has a mass of 32.0 𝑘𝑔, 𝑟2 = 𝑟 𝜏𝑝 = 𝑟𝑝 𝐹𝑝 = 0. 𝐹𝑝 = 0 𝑚2 1 how far is she from the pivot? b) What is 𝐹⊥ , the supporting force exerted by 𝑚1 26 𝑟2 = 𝑟1 = 1.6 = 1.30 𝑚 the pivot using the second condition for 𝑚2 32 equilibrium 𝑛𝑒𝑡 𝜏 = 0 , employing any data given or solved for in part (a).​ (b) 𝑛𝑒𝑡 𝐹 = 0 𝑎𝑛𝑑 𝐹𝑦 = 0 𝐹𝑝 − 𝑊1 − 𝑊2 = 0 𝐹𝑝 = 𝑊1 + 𝑊2 = 𝑚1 𝑔 + 𝑚2 𝑔 𝐹𝑝 = 26 9.8 + 32 9.8 = 568 𝑁 32  Activities…. Exercise: Two children seat themselves on a Example: A 5 𝑚, 10 𝑘𝑔 seesaw is balanced by a seesaw. The one on the left has a weight of 400 little girl 25 𝑘𝑔 and her father 80 𝑘𝑔 at N while the one on the right weighs 300 N. The opposite ends as shown below. fulcrum is at the midpoint of the seesaw. If the a) How far from the seesaw’s centre of mass child on the left is not at the end but is 1.50 m must the fulcrum be placed? from the fulcrum and the seesaw is balanced, b) How much force must the fulcrum support? what is the torque provided by the weight of the child on the right? (a) 𝜏𝐶𝑊 = 𝜏𝐶𝐶𝑊 80 × 9.8 × 𝑥 = 25 × 9.8 × 5 − 𝑥 𝑥 = 1.19 𝑚 33  Activities…. (b) 𝐹𝑓𝑢𝑙𝑐𝑟𝑢𝑚 − 25 × 9.8 − 80 × 9.8 = 0 Exercise: Consider a 75 𝑘𝑔 mother and her 𝐹𝑓𝑢𝑙𝑐𝑟𝑢𝑚 = 1029 𝑁 15 𝑘𝑔 daughter, balancing on a seesaw, which is a uniform board 3 𝑚 long and pivoted Exercise: The seesaw is 4.5 m long. Its mass of 20 kg exactly at its center. If the daughter sits at the is uniformly distributed. The child on the left end has end of the seesaw (1.5 𝑚 from the center), a mass of 14 kg and is a distance of 1.4 m from the how far from the center on the other side pivot point while a second child of mass 39 kg stands should the mother sit? a distance L from the pivot point, keeping the seesaw at rest. Find L the distance of the child on the right? 34 Engineering statics 35  Engineering Statics Engineering statics is the study of rigid For an object in equilibrium 𝑎 = 0 which bodies in equilibrium so it’s appropriate to implies that the body is either stationary or begin by defining what we mean by rigid moving with a constant velocity bodies and what we mean by equilibrium. 𝑣=0 𝑎=0⇒ቊ 𝑣=𝐶 A rigid body is a body that doesn’t deform The acceleration of an object is related to the under load, that is to say, an object which net force acting on it by Newton’s Second doesn’t bend, stretch, or twist when forces are Law applied to it. ෍ 𝐹 = 𝑚𝑎 A body in equilibrium is not accelerating. As So, for the special case of static equilibrium you learned in physics, acceleration is Newton’s Law becomes velocity’s time rate of change and is a vector quantity. For linear motion, ෍𝐹 = 0 𝑑𝑣 𝑎= If a body is in equilibrium, then all the forces 𝑑𝑡 acting on the body must sum to zero. 36  Particles The defining characteristic of a particle is Particles in One Dimension that all forces that act on it are coincident In mechanics we are interested in studying or concurrent. the forces acting on objects. Forces are coincident if they have the A simple case same line of action. Forces are concurrent if they intersect The free-body diagram: A diagram which at a point. focuses on the forces acting on an object, not All forces will be assumed to be the mechanisms that hold it in place. concentrated. Concentrated forces act at a single point, have a well-defined line of action, and can be represented with an arrow — in other words, they are vectors. 37  …. Scalar Components Drawing free-body diagrams can be The scalar component of a vector is a signed surprisingly tricky. number which indicates the vector’s Must identify all the forces acting on the magnitude and sense. object and correctly represent them on Scalar components can be added together the free-body diagram. algebraically, but only if they act “in the same If not, all forces are accounted for or direction. they are represented them incorrectly, Example: If 𝐹𝑥 = −40 𝑁 and 𝐹𝑦 = 30 𝑁, find the the analysis will be incorrect. magnitude and direction of their resultant. Principle of Transmissibility: A force can be moved along it’s line of action and the net external effects remain the same. 38  … ❖ A two-force body is a body with two forces acting on it. ❖ The two forces must either Share the same line of action, have the same magnitude, and point away from each other, or Particles in Two Dimensions Share the same line of action, have the The general procedure for solving equilibrium same magnitude, and point towards each of a particle problems in two dimensions is to: other, or Both have zero magnitude. When two forces have the same magnitude but act in diametrically opposite directions, we say that they are equal-and-opposite. Compression - point towards each other. Tension - point away from each other. 39  … Force Triangle Method Trigonometric Method ❖ Applicable to situations where there are (exactly) three forces acting on a particle, The general approach for solving particle and no more than two unknown magnitudes equilibrium problems using the or directions. trigonometric method is to: ▪ If such a particle is in equilibrium, then the 1) Draw and label a free-body diagram. three forces must add to zero. 2) Rearrange the forces into a force triangle ▪ Graphically, force vectors are arranged tip- and label it. to-tail, and form a closed, three-sided 3) Identify the knowns and unknowns polygon. (there must be no more than two unknowns). 4) Use trigonometry to find the unknown sides or angles of the triangle. 40  … LAMI’S THEOREM Quiz: Differentiate translational from rotational equilibrium (3 marks). If a system of three forces is in equilibrium, T: The linear speed is not changing with time. There is no then, each force of the system is proportional resultant force and therefore zero acceleration. to sine of the angle between the other two R: The angular speed is not changing with time. There is forces (and constant of proportionality is the no resultant torque and, therefore, zero change in rotational velocity. same for all the forces). Example: Find the tension in ropes A and B. 𝐹1 𝐹2 𝐹3 = = 𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝛽 𝑠𝑖𝑛𝛾 Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence. 41  …. We have three forces: 𝑇𝐵 80 𝑇𝐴 (horizontal, along cable A), = sin 90 sin 120 𝑇𝐵 (at a 60° angle to the horizontal, along cable B), 80 𝑊=80N (vertical downward force). 𝑇𝐵 = = 92.4 𝑁 sin 120 𝑇𝐴 𝑇𝐵 𝑊 = = 𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝛽 𝑠𝑖𝑛𝛾 𝑇𝐴 𝑇𝐵 𝛼 is the angle opposite to 𝑇𝐴 ​ , which is 60° = sin 60 sin 90 (between 𝑊 and 𝑇𝐵 ​) 𝛽 is the angle opposite to 𝑇𝐵 ​, which is 90° (between 𝑇𝐴 ​ and 𝑊), 𝑇𝐵 sin 60 𝑇𝐴 = = 46.2 𝑁 𝛾 is the angle opposite to 𝑊, which is 120° sin 90 (between 𝑇𝐴 and 𝑇𝐵 ). 𝑇𝐴 𝑇𝐵 80 = = sin 60 sin 90 sin 120 42  … Example: Find tension in ropes A and B. 𝑇𝐴 𝑇𝐵 500 = = sin 55𝑂 sin 35𝑂 sin 90𝑂 𝑇𝐴 500 𝑂 = sin 55 sin 90𝑂 𝑇𝐴 = 500 sin 55𝑂 = 409.5 𝑁 𝑇𝐴 ​ and 𝑇𝐵 ​ ​ are the tensions in cables A and B, respectively. 𝑇𝐵 500 𝑂 = 𝑊 = 500 𝑁 is the weight acting vertically sin 35 sin 90𝑂 downwards. 𝛼 = 55𝑂 , which is the angle opposite 𝑇𝐴 𝛽 = 35𝑂 , which is the angle opposite 𝑇𝐵 𝑇𝐵 = 500 sin 35𝑂 = 286.8 𝑁 𝛾 = 180𝑂 − 35𝑂 + 55𝑂 = 90𝑂 , which is the angle opposite the weight 𝑊. 43  Scalar Components Method The general statement of equilibrium of Example: Find the tension in ropes A and B. forces, can be expressed as the sum of forces ෡ directions. in the 𝒊,Ƹ 𝒋Ƹ and 𝒌 ෍ 𝐹 = ෍ 𝐹𝑥 𝑖Ƹ + ෍ 𝐹𝑦 𝑗Ƹ + ෍ 𝐹𝑧 𝑘෠ = 0 This statement will only be true if all three coefficients of the unit vectors are themselves equal to zero, three dimensions The weight force 𝑊 = 80𝑁 acting ෍ 𝐹𝑥 = 0 vertically downward. The tension in cable A 𝑇𝐴 acting ෍ 𝐹 = 0 ⇒ ෍ 𝐹𝑦 = 0 (three dimensions) horizontally. The tension in cable B 𝑇𝐵 acting at a 60° ෍ 𝐹𝑧 = 0 angle to the horizontal. 44  Examples… Example: A long rope is stretched between points A Resolve forces into components: Since the and B. At each end, the rope is tied to a spring scale that system is in static equilibrium, the sum of measures the force the rope exerts on the supports. the forces in both the x (horizontal) and y Suppose the rope is pulled sideways at its midpoint with (vertical) directions must be zero. a force of 400 𝑁 producing a deflection such that the Vertical forces two segments make angles of 5° with the line AB. What 𝑇𝐵 sin 60 = 80 𝑁 is the reading of the spring scales? 80 𝑇𝐵 = = 92.4 𝑁 sin 60 X-components Horizontal forces 𝑇𝐴 cos 5 − 𝑇𝐵 cos 5 = 0 (𝑖) Y-components 𝑇𝐴 = 𝑇𝐵 cos 60 𝑇𝐴 sin 5 + 𝑇𝐵 sin 5 − 400 = 0 (𝑖𝑖) From 𝑖 , 𝑇𝐴 = 𝑇𝐵 𝑇𝐴 = 𝑇𝐵 cos 60 = 92.4 × 0.5 = 46.2 𝑁 𝑇𝐵 sin 5 + 𝑇𝐵 sin 5 − 400 = 2𝑇𝐵 sin 5 − 400 (𝑖𝑖𝑖) 400 𝑇𝐵 = = 2295 = 𝑇𝐴 2 sin 5 45  Examples… Example: A 2-kg ball is held in position by a ෍ 𝐹𝑥 = −𝑇 + 𝑇𝑟 cos 60 = 0 horizontal string and a string that makes an angle of 30° with the vertical, as shown in the −𝑇 + 𝑇𝑟 0.5 = 0 figure. Find the tension T in the horizontal string. 𝑇 = 𝑇𝑟 (0.5) ෍ 𝐹𝑦 = 𝑇𝑟 sin 60 − 20 𝑁 = 0 𝑇𝑟 0.866 = 20 𝑁 20 𝑇𝑟 = = 23.1 𝑁 0.866 ∴ 𝑇 = 23.1 0.5 = 11.5 𝑁 46  Activities Example: The three forces in the diagram Exercise: Determine if the forces are at are in equilibrium. What are the values of B equilibrium. and θ? Exercise: The diagram shows an object, of mass 300 kg, that is at rest and is supported by two cables. Find the tension in each cable. 47  Activities Exercise: A particle of mass 6 kg is suspended Exercise: A child, of mass 30 kg, slides down a slide by two strings as shown in the diagram. Note at a constant speed. Assume that there is no air that one string is horizontal. Find the tension in resistance acting on the child. The slide makes an each string. angle of 40o with the horizontal. Find the magnitude of the friction force on the child and the coefficient of friction. Exercise: A lorry of mass 5000 kg drives up a slope inclined at 5o to the horizontal. The lorry moves in a straight line and at a constant speed. Exercise: A crate of mass 200 kg is on Assume that no resistance forces act on the a horizontal surface. The coefficient of lorry. Find the magnitude of the normal reaction friction between the crate and the force and force that acts on the lorry in its surface is 0.4. The crate is pulled by a direction of motion. rope, as shown in the diagram, so that the crate moves at a constant speed in a straight line. Find the tension in the rope. 48