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2:1 MM203Dr. Alan Kennedy MM203 Mechanics of Machines: Part 2

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2:2 MM203Dr. Alan Kennedy Kinetics of systems of particles Extension of basic principles to general systems of particles –Particles with light links –Rigid bodies –Rigid bodies with flexible links –Non-rigid bodies –Masses of fluid

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2:3 MM203Dr. Alan Kennedy Newtons second law G – centre of mass F i – external force, f i – internal force i – position of m i relative to G

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2:4 MM203Dr. Alan Kennedy Newtons second law By definition For particle i Adding equations for all particles

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2:5 MM203Dr. Alan Kennedy Newtons second law Differentiating w.r.t. time gives Also so (principle of motion of the mass centre)

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2:6 MM203Dr. Alan Kennedy Newtons second law Note that is the acceleration of the instantaneous mass centre – which may vary over time if body not rigid. Note that the sum of forces is in the same direction as the acceleration of the mass centre but does not necessarily pass through the mass centre

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2:7 MM203Dr. Alan Kennedy Example Three people ( A, 60 kg, B, 90 kg, and C, 80 kg) are in a boat which glides through the water with negligible resistance with a speed of 1 knot. If the people change position as shown in the second figure, find the position of the boat relative to where it would be if they had not moved. Does the sequence or timing of the change in positions affect the final result? (Answer: x = m ). (Problem 4/15, M&K)

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2:8 MM203Dr. Alan Kennedy Example The 1650 kg car has its mass centre at G. Calculate the normal forces at A and B between the road and the front and rear pairs of wheels under the conditions of maximum acceleration. The mass of the wheels is small compared with the total mass of the car. The coefficient of static friction between the road and the rear driving wheels is 0.8. (Answer: N A = 6.85 kN, N B = 9.34 kN ). (Problem 6/5, M&K)

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2:9 MM203Dr. Alan Kennedy Work-energy Work-energy relationship for mass i is where (U 1-2 ) i is the work done on m i during a period of motion by the external and internal forces acting on it. Kinetic energy of mass i is

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2:10 MM203Dr. Alan Kennedy Work-energy For entire system

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2:11 MM203Dr. Alan Kennedy Work-energy Note that no net work is done by internal forces. If changes in potential energy possible (gravitational and elastic) then as for single particle

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2:12 MM203Dr. Alan Kennedy Work-energy For system Now and note that so

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2:13 MM203Dr. Alan Kennedy Work-energy Since i is measured from G, Now

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2:14 MM203Dr. Alan Kennedy Work-energy Therefore i.e. energy is that of translation of mass-centre and that of translation of particles relative to mass-centre

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2:15 MM203Dr. Alan Kennedy Example The two small spheres, each of mass m, are rigidly connected by a rod of negligible mass and are released in the position shown and slide down the smooth circular guide in the vertical plane. Determine their common velocity v as they reach the horizontal dashed position. Also find the force R between sphere 1 and the guide the instant before the sphere reaches position A. (Answer: v = 1.137(gr) ½, R = 2.29mg ). (Problem 4/9, M&K)

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2:16 MM203Dr. Alan Kennedy Rigid body Motion of particles relative to mass-centre can only be due to rotation of body Velocity of particles due to rotation depends on angular velocity and the distance to centre of rotation. Where is centre of rotation? Need to examine kinematics of rotation

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2:17 MM203Dr. Alan Kennedy Plane kinematics of rigid bodies Rigid body –distances between points remains unchanged –position vectors, as measured relative to coordinate system fixed to body, remain constant Plane motion –motion of all points is on parallel planes –Plane of motion taken as plane containing mass centre –Body treated as thin slab in plane of motion – all points on body projected onto plane

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2:18 MM203Dr. Alan Kennedy Kinematics of rigid bodies

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2:19 MM203Dr. Alan Kennedy Translation All points move in parallel lines or along congruent curves. Motion is completely specified by motion of any point – therefore can be treated as particle Analysis as developed for particle motion

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2:20 MM203Dr. Alan Kennedy Kinematics of rigid bodies

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2:21 MM203Dr. Alan Kennedy Rotation about fixed axis All particles move in circular paths about axis of rotation All lines on body (in plane of motion) rotate through the same angle in the same time Similar to circular motion of a particle where r iO is distance to O, the centre of rotation, and I O is mass moment of inertia about O

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2:22 MM203Dr. Alan Kennedy Mass moment of inertia Mass moment of inertia in rotation is equivalent to mass in translation Rotation and translation are analogous

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2:23 MM203Dr. Alan Kennedy General plane motion Combination of translation and rotation Principles of relative motion used

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2:24 MM203Dr. Alan Kennedy Rotation Angular positions of two lines on body are measured from any fixed reference direction

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2:25 MM203Dr. Alan Kennedy Rotation All lines on a rigid body in its plane of motion have the same angular displacement, the same angular velocity, and the same angular acceleration Angular motion does not require the presence of a fixed axis about which the body rotates

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2:26 MM203Dr. Alan Kennedy Angular motion relations Angular position, angular velocity, and angular acceleration Similar to relationships between s, v, and a. Also, combining relationships and cancelling out dt

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2:27 MM203Dr. Alan Kennedy Angular motion relations If constant angular acceleration Direction of +ve sense must be consistent Analogous to rectilinear motion with constant a Same procedures used in analysis

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2:28 MM203Dr. Alan Kennedy Example The angular velocity of a gear is controlled according to = 12 – 3t 2 where, in radians per second, is positive in the clockwise sense and where t is the time in seconds. Find the net angular displacement from the time t = 0 to t = 3 s. Also find the total number of revolutions N through which the gear turns during the 3 seconds. (Answer: = 9 rad, N = 3.66 rev ). (Problem 5/5, M&K)

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2:29 MM203Dr. Alan Kennedy Kinetic energy of rigid body If rotation about O

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2:30 MM203Dr. Alan Kennedy Parallel axis theorem Now (P.A.T.) and so

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2:31 MM203Dr. Alan Kennedy Radius of gyration Mass moment of inertia of point mass m at radius of gyration is the same as that for body P.A.T.

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2:32 MM203Dr. Alan Kennedy Work done on rigid body

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2:33 MM203Dr. Alan Kennedy Work done by couple Couple is system of forces that causes rotation but no translation Moment about G Moment about O

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2:34 MM203Dr. Alan Kennedy Work done by couple Moment vector is a free vector Forces have turning effect or torque Torque is force by perpendicular distance between forces Work done positive or negative

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2:35 MM203Dr. Alan Kennedy Forces and couples Torque is Also unbalanced force

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2:36 MM203Dr. Alan Kennedy Work-energy principle When applied to system of connected bodies only consider forces/moments of system – ignore internal forces/moments. If there is significant friction between components then system must be dismembered

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2:37 MM203Dr. Alan Kennedy Example A steady 22 N force is applied normal to the handle of the hand- operated grinder. The gear inside the housing with its shaft and attached handle have a combined mass of 1.8 kg and a radius of gyration about their axis of 72 mm. The grinding wheel with its attached shaft and pinion (inside housing) have a combined mass of 0.55 kg and a radius of gyration of 54 mm. If the gear ratio between gear and pinion is 4:1, calculate the speed of the grinding wheel after 6 complete revolutions of the handle starting from rest. (Answer: N = 3320 rev/min ). (Problem 6/119, M&K)

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2:38 MM203Dr. Alan Kennedy Rotation about fixed axis Motion of point on rigid body

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2:39 MM203Dr. Alan Kennedy Vector notation Angular velocity vector,, for body has sense governed by right-hand rule free vector

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2:40 MM203Dr. Alan Kennedy Vector notation Velocity vector of point A What are magnitude and direction of this vector? Note that

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2:41 MM203Dr. Alan Kennedy Vector notation Acceleration of point

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2:42 MM203Dr. Alan Kennedy Vector notation Vector equivalents Can be applied in 3D except then angular velocity can change direction and magnitude

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2:43 MM203Dr. Alan Kennedy Example The T-shaped body rotates about a horizontal axis through O. At the instant represented, its angular velocity is = 3 rad/s and its angular acceleration is = 14 rad/s 2. Determine the velocity and acceleration of (a) point A and (b) point B. Express your results in terms of components along the n - and t - axes shown. (Answer: v A = 1.2e t m/s, a A = 5.6e t + 3.6e n m/s 2, v B = 1.2e t + 0.3e n m/s, a B = 6.5e t + 2.2e n m/s 2 ). (Problem 5/2, M&K)

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2:44 MM203Dr. Alan Kennedy Example The two V-belt pulleys form an integral unit and rotate about the fixed axis at O. At a certain instant, point A on the belt of the smaller pulley has a velocity v A = 1.5 m/s, and the point B on the belt of the larger pulley has an acceleration a B = 45 m/s 2 as shown. For this instant determine the magnitude of the acceleration a C of the point C and sketch the vector in your solution. (Answer: a C = m/s 2 ). (Problem 5/16, M&K)

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2:45 MM203Dr. Alan Kennedy Linear impulse and momentum Returning to general system: Linear momentum of mass i is For system (assuming m does not change with time)

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2:46 MM203Dr. Alan Kennedy Linear impulse and momentum Differentiating w.r.t. time Same as for single particle – only applies if mass constant Same for rigid body

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2:47 MM203Dr. Alan Kennedy Example The 300 kg and 400 kg mine cars are rolling in opposite directions along a horizontal track with the speeds shown. Upon impact the cars become coupled together. Just prior to impact, a 100 kg boulder leaves the delivery chute and lands in the 300 kg car. Calculate the velocity v of the system after the boulder has come to a rest relative to the car. Would the final velocity be the same if the cars were coupled before the boulder dropped? (Answer: v = m/s ). (Problem 4/11, M&K)

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2:48 MM203Dr. Alan Kennedy Angular impulse and momentum Considered about a fixed point O and about the mass centre.

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2:49 MM203Dr. Alan Kennedy Angular impulse and momentum About O First term is zero since v i × v i =0 so - sum of all external moments (net moment of internal forces is zero)

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2:50 MM203Dr. Alan Kennedy Angular impulse and momentum About O : same as for single particle. As before, does not apply if mass is changing. About G

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2:51 MM203Dr. Alan Kennedy Example The two balls are attached to a light rod which is suspended by a cord from the support above it. If the balls and rod, initially at rest, are struck by a force F = 60 N, calculate the corresponding acceleration aˉ of the mass centre and the rate d 2 /dt 2 at which the angular velocity of the bar is changing. (Answer: aˉ = 20 m/s 2, d 2 /dt 2 = 336 rad/s 2 ). (Problem 4/17, M&K)

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2:52 MM203Dr. Alan Kennedy Rigid body Angular momentum Planar motion

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2:53 MM203Dr. Alan Kennedy Rigid body Angular acceleration

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2:54 MM203Dr. Alan Kennedy Kinetic diagrams

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2:55 MM203Dr. Alan Kennedy Kinetic diagrams - translation Alternative moment equation for rectilinear translation

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2:56 MM203Dr. Alan Kennedy Kinetic diagrams - translation Alternative moment equation for curvilinear translation

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2:57 MM203Dr. Alan Kennedy Example The cart B moves to the right with acceleration a = 2g. If the steady-state angular deflection of the uniform slender rod of mass 3 m is observed to be 20°, determine the value of the torsional spring constant K. The spring, which exerts a moment M = K on the rod, is undeformed when the rod is vertical. The values of m and l are 0.5 kg and 0.6 m, respectively. Treat the small end sphere of mass m as a particle. (Answer: K = 46.8 N·m/rad ). (Problem 6/16, M&K)

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2:58 MM203Dr. Alan Kennedy Example The mass of gear A is 20 kg and its centroidal radius of gyration is 150 mm. The mass of gear B is 10 kg and its centroidal radius of gyration is 100 mm. Calculate the angular acceleration of gear B when a torque of 12 N·m is applied to the shaft of gear A. Neglect friction. (Answer: B = 25.5 rad/s 2 (CCW) ). (Problem 6/46, M&K)

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2:59 MM203Dr. Alan Kennedy Example The 28 g bullet has a horizontal velocity of 500 m/s when it strikes the 25 kg compound pendulum, which has a radius of gyration of k O = 925 mm. If the distance h = 1075 mm, calculate the angular velocity of the pendulum with its embedded bullet immediately after the impact. (Answer: = rad/s) (Problem 6/174, M&K)

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2:60 MM203Dr. Alan Kennedy Plane kinematics: absolute motion Absolute motion analysis –Get geometric relationships –Get time derivatives to determine velocity and acceleration –Straightforward if geometry is straightforward –Must be consistent with signs

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2:61 MM203Dr. Alan Kennedy Example Point A is given a constant acceleration a to the right starting from rest with x essentially 0. Determine the angular velocity of link AB in terms of x and a. (Problem 5/24, M&K) Answer:

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2:62 MM203Dr. Alan Kennedy Example The wheel of radius r rolls without slipping, and its centre O has a constant velocity v O to the right. Determine expressions for the velocity v and acceleration of point A on the rim by differentiating its x - and y - coordinates. Represent your result graphically as vectors on your sketch and show that v is the vector sum of two v O vectors. (Problem 5/25, M&K) Answer:

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2:63 MM203Dr. Alan Kennedy Example One of the most common mechanisms is the slider-crank. Express the angular velocity AB and the angular acceleration AB of the connecting rod AB in terms of the crank angle for a given constant crank speed 0. Take AB and AB to be positive counter-clockwise. (Problem 5/54, M&K) Answer:

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2:64 MM203Dr. Alan Kennedy Plane kinematics: relative velocity Two points on same rigid body. Motion of one relative to the other must be circular since distance between them is constant.

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2:65 MM203Dr. Alan Kennedy Plane kinematics: relative velocity Relative linear velocity is always in direction perpendicular to line joining points

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2:66 MM203Dr. Alan Kennedy Plane kinematics: relative velocity

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2:67 MM203Dr. Alan Kennedy Plane kinematics: relative velocity Relative velocity principles may also be used in cases where there is constrained sliding contact between two links – A and B may be on different links

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2:68 MM203Dr. Alan Kennedy Example Determine the angular velocity of the telescoping link AB for the position shown where the driving links have the angular velocities indicated. (Answer: AB = 0.96 rad/s ). (Problem 5/61, M&K)

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2:69 MM203Dr. Alan Kennedy Example For an interval of its motion the piston rod of the hydraulic cylinder has a velocity V A = 1.2 m/s as shown. At a certain instant = = 60°. For this instant determine the angular velocity BC of the link BC. (Answer: BC = rad/s). (Problem 5/66, M&K)

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2:70 MM203Dr. Alan Kennedy Example The mechanism is designed to convert from one rotation to another. Rotation of link BC is controlled by the rotation of the curved slotted arm OA which engages pin P. For the instant represented = 30° and, the angle between the tangent to the curve at P and the horizontal, is 40°. If the angular velocity of OA is as shown, determine the velocity of the point C. (Answer: v C = 4.33 m/s). (Problem 5/85, M&K)

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2:71 MM203Dr. Alan Kennedy Relative acceleration May need to know velocities first

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2:72 MM203Dr. Alan Kennedy Example If OA has a CCW angular velocity 0 = 10 rad/s (giving BC = 5.83 rad/s and AB = 2.5 rad/s ), calculate the angular acceleration of link AB for the position where the coordinates of A are x =60 mm and y = 80 mm. Link BC is vertical for this position. (Answer: AB = 2.5 rad/s 2 ). (Problem 5/137, M&K)

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