Presentation on theme: "ENGINEERING MECHANICS CHAPTER 7"— Presentation transcript:
1ENGINEERING MECHANICS CHAPTER 7 School of EngineeringENGINEERING MECHANICSCHAPTER 7Kinematics of Linear & RotationalMotionClick on VIEW and select SLIDE SHOW to view the presentation.
26.1) Introduction To Kinematics of Linear Motion Kinematics is the analysis of the geometry of motion without considering the forces causing the motion. It involves quantities like displacement (m), velocity (m/s), acceleration (m/s2) and time (s).6.1.1 DisplacementDisplacement is defined as the location of the body with respect to a reference point. It specifies the distance and the direction from the reference point. The SI unit for displacement is m.
3Displacement-time graph is a plot of the displacement against time. Example:An object is thrown vertically upward from the top of a building. The starting location is taken as the reference point. Its displacement from starting location is as shown in the figure below.Negative displacementstMaximum heightZero displacementReferencepoint
46.1.2 VelocityVelocity is the rate of change of displacement with respect to time. The SI unit for velocity is m/s.v = ds/dtVelocity-time GraphThe velocity-time graph is useful to visualize the motion. Using the same example above, the plot of the velocity with respect to time is as shown in the figure below. During the upward journey, the velocity is positive. It decreases to zero when the body reaches the maximum height. The velocity becomes negative on the downward journey.
5Zero velocity at maximum height Positive velocityNegative velocityAccelerationAcceleration is the rate of change of velocity with respect to time. The SI unit for acceleration is m/s2.a = dv/dtUsing the same example as above, the acceleration will be a constant and is negative for the whole journey. The magnitude of the acceleration is 9.81 m/s2 for free falling motion.
6Uniform motionA motion is called uniform motion when the acceleration is constant.The three equations of motion with constant acceleration are:v = vi + atv2 = vi2 + 2ass = vit + ½at2where v is the final velocityvi is the initial velocitya is the accelerations is the displacementt is the time
7Example 6.1A projectile is ejected vertically from ground level upward with an initial velocity of 30 m/s. Determine:i) its velocity after 2 s,ii) the height above the ground at this time,iii) the time taken to reach its maximum height,iv) the total time taken for it to hit the ground again.Solution:i) Take upward as positive direction.vi = 30 m/sa = m/s2t = 2 si.e v = vi + at= 30 + (- 9.81) x 2= m/s
8ii) s = vit + ½at2= 30 x 2 + ½ x (-9.81) x 22= miii) At maximum height, velocity is zero.v = vi + at0 = 30 + (-9.81) x tt = siv) s = vit + ½at20 = 30 x t + ½ x (-9.81) x t2t = 0 or s
9Non-uniform MotionA moving body with non-uniform motion has different accelerations (and/or decelerations) during the whole motion.The motion can be divided into several segments with constant acceleration for each segment.The three equations for constant acceleration can then be applied for each segment where the motion is uniform.
10i) the velocity at point B, ii) the distance from A to B, Example 6.2An MRT train travels between two stations A and D. It starts with an acceleration of 1.2 m/s2 for 18 s until it reaches point B. The velocity remains constant over a distance of 2000 m from point B to C, and then it decelerates uniformly at 2.4 m/s2 until it stops at D. Determine:i) the velocity at point B,ii) the distance from A to B,iii) the distance C to D,iv) the total time and distance.DCBAvtSolution:Velocity - Time Graph
11i) From A to B, vi = 0 m/s a = 1.2 m/s2 t = 18 s, v = vi + at DCBAvt- 2.4 m/s21.2 m/s218 st2t1i) From A to B,vi = 0 m/sa = 1.2 m/s2t = 18 s, v = vi + at= x 18= 21.6 m/sii) From A to B,s = vit + ½at2= 0 x 18 + ½ x 1.2 x 182= miii) From C to D, vi = 21.6 m/sAlso, v = 0 and a = m/s2v2 = vi2 + 2as0 = x (-2.4) x ss = 97.2 m
12iv) Total distance, sT = 194.4 + 2000 + 97.2 = 2291.6 m BAvt- 2.4 m/s21.2 m/s218 st2t1iv) Total distance,sT == mFrom B to C, vi= 21.6 m/sa = 0, s = 2000 ms = vit + ½at22000 = 21.6 x t + 0t = s (t1)From C to D,v = vi + at0 = (-2.5) x tt = 8.64 s (t2)Total time, tT == s (Ans)
136.2) Introduction To Kinematics of Rotational Motion Kinematics of rotational motion involves quantities like angular displacement, angular velocity, angular acceleration and time.Angular DisplacementAngular displacement is defined as the angle and the direction through which a body turns. The SI unit for angular displacement is radian (rad).( Note:- 1 revolution = 360 degrees = 2 radians )Angular VelocityAngular velocity is the rate of change of angular displacement with respect to time. The angularvelocity can be written in mathematical form as: = d/dtThe SI unit for angular velocity is rad/s.
14Angular acceleration is the rate of change of angular velocity with respect to time. The angularacceleration can be written in mathematical form as: = d/dtThe SI unit for is rad/s2.Uniform motionA motion is called uniform when the angular acceleration is constant.The three equations of rotational motion with constant acceleration are:Note the similarities of these equations with thoseof linear motion in above.Where is the final velocityi is the initial velocity is the acceleration is the displacementt is the time = i + t2 = i2 + 2 = it + ½t2
156.2.5 Relationship Between Linear and Rotational motion The equations for the relationship between linear and rotational motion are:s = rv = r ( Divide s = r by t )a = r ( Divide v = r by t )where r is the radius of rotationThe linear velocity and acceleration are also the tangential velocity and acceleration respectively for a point on the circle of rotation.( Note: v = vt and a = at )s = r
16Example 6.3A propeller fan used in a cooling tower comes to rest with uniform deceleration from a speed of 600 rpm. It turns through 15 rev while stopping. Determinethe initial blade tip speed if the blade is 2 m long,the deceleration,the time taken to stop.Solutioni. i = 600 rpm= (600 x 2) / 60= rad/sv = r= 2 x 62.83= m/s
17ii = 15 rev = 15 x 2= 30 rad = 0 , i =62.83 rad/ssince 2 = i2 + 20 = x x 30 = rad/s2iii = i + t0 = ( )tt = 3 s
18the number of revolutions made by the pulley in 3 s, Example 6.4A compound pulley has an outer radius of 1.25 m and an inner radius of 0.75 m. Load A and load B are connected by cords to the pulley at the outer radius and inner radius respectively. Load A has a constant acceleration of 2.5 m/s2 and an initial velocity of 3.75 m/s, both in the upward direction. Determinethe number of revolutions made by the pulley in 3 s,the velocity of load B in this time,the distance moved by load B.ABr = 0.75 mr = 1.25 mvi = 3.75 m/sa = 2.5 m/s2
19i) vi = ri 3.75 = 1.25i i = 3 rad/s a = r 2.5 = 1.25 = 2 rad/s2 Br = 0.75 mr = 1.25 mvi = 3.75 m/sa = 2.5 m/s2At t = 3 s, = it + ½t2= 3 x 3 + ½ x 2 x 32= 18 radNumber of revolutions,N = 18 / 2= 2.87 rev
20iii) Distance moved by load B, s = r = 0.75 x 18 = 13.5 m ii) = i + t= x 3= 9 rad/sv = r= 0.75 x 9= 6.75 m/sABr = 0.75 mr = 1.25 mvi = 3.75 m/sa = 2.5 m/s2iii) Distance moved by load B,s = r= 0.75 x 18= 13.5 m
216.2.6 Normal or Centripetal Acceleration For a body moving in a circular motion, it has an acceleration an, towards the center of the circle with a magnitude given by,at (TangentialAcceleration)(Normal or Centripetal Acceleration) anaT (Total Acceleration)an = vor an = v2/r (Since = v/r)or an = r (Since v = r)( Note:- Total acceleration aT2 = an2 + at2 )
22Example 6.5Disk A and disk B are on the same plane. Disk A starts from rest and rotates with a constant angular acceleration of 2 rad/s2. It is in contact with disk B and no slipping occurs between them. Determine the angular velocity and angular acceleration of B just after A turns 10 revolutions. What is the total acceleration of the point P in relation to disk B?ABP2m1.5m
23Velocity of point P on disk A is vA = rAA = 2 x 15.85 = 31.7 m/s SolutionA2 = Ai2 + 2A= x 2 x (10 x 2)A = rad/sVelocity of point P on disk A isvA = rAA= 2 x 15.85= 31.7 m/sABP2m1.5mSince there is no slipping, vB = vA = 31.7 m/sThen vB = rBB31.7 = 1.5BB = rad/sAcceleration of the point P on disk A isaA = rAA = 2 x 2= 4 m/s2
24End of Chapter 6 Since there is no slipping, aB = aA = 4 m/s2 Then aB = rBB4 = 1.5BB = 2.67 rad/s2ABP2m1.5mFor point P in relation to disk B, at = aB = 4 m/s2an = r2= 1.5 x= m/s2aT2 = at2 + an2=aT = m/s2End of Chapter 6