Revision of Linear and Angular Displacement,, and Acceleration The following symbols are used in these notes: Linear Terms Symbol Units Displacement x or s m Initial velocity u m/s Final velocity v m/s Acceleration a m/s 2 Time t s Angular Terms Angular displacement θ rad Initial angular velocity rad/s Final angular velocity rad/s Angular acceleration rad/s 2 Linear Motion The velocity of a body is defined as the rate of change of position with time. If the body travels distance s meters in time t seconds then the average velocity is given by: m/s Note that velocity has magnitude and direction and is therefore a vector quantity (see below). θ v Speed on the other hand is merely the magnitude of its velocity and is therefore a scalar quantity. Acceleration Acceleration of a body is defined as the rate of change of velocity with time. A body therefore accelerates if there is a change in its magnitude and/or direction and/or sense of its velocity. Note that acceleration can therefore occur without a change in speed. For example a body moving in a circular path at uniform speed, the direction of its tangential velocity is constantly changing. Equations of Motion The following relationships can be derived: 1
It should be noted that increasing speed means that the acceleration is positive; decreasing speed means that the acceleration is negative. -Time Graphs A typical velocity-time graph is given below. Provided the accelerations and decelerations are uniform the graph will be a series of straight lines. The area under the graph represents the total distance travelled. Area = Distance travelled Time Worked Example: (From Applied Mechanics by Hannah & Hillier) The cutting stroke of a planning machine is 600 mm and it is completed in 1.2 s. For the first and last quarter of the stroke the table is uniformly accelerated and retarded, the speed remains constant for the remainder of the stroke. Using a speed-time graph, or otherwise, determine the maximum cutting speed. (0.75 m/s) Solution: This can be represented on a velocity-time graph as follows: v A 2 A 1 A 3, t 1 t 2 Time Distance travelled during acceleration and deceleration m Therefore m and m Now, therefore --------------(1) But -------------(2) But ---------------(3) Substituting (3) into (2) gives 2
-------------(4) Substituting (1) into (4) gives Therefore or s Substituting this back into (1) gives v m/s = Answer Tutorial Problems (From Applied Mechanics by Hannah & Hillier 1. A diesel train accelerates uniformly from rest to reach 60 km/h in 6 mins, after which the speed is kept constant. Calculate the total time taken to travel 6 kilometres. (9 mins) 2. The driver of a train shuts off the power and the train is uniformly retarded. In the first 30 s the train covers 110 m, and then comes to rest in a further 30s. Determine (a) the initial speed of the train before the power is cut off, (b) the total distance travelled in coming to rest. (4.89 m/s; 147 m) Freely Falling Objects When a object falls freely to earth, it accelerates uniformly downwards with a acceleration g. Close to the earth the value of g is found to be approximately equal to 9.8 m/s 2. Similarly if an object is projected upwards it decelerates at a uniform rate also equal to g. In both cases this assumes that air resistance is neglected. The equation for uniformly accelerated motion in a straight line also apply to bodies that are freely falling or moving upwards freely the proviso being that a is replaced by g (positive or negative depending whether the object is falling or moving upwards). Worked Example A space rocket is launched vertically from rest and reaches a height h. The acceleration is 10.5 m/s 2 until its fuel burns out after 6.6 s. Assuming, at this point that the rocket continues to travel freely vertically upwards with constant deceleration g = 9.8 m/s 2, determine the height h. The solution has two parts to it; the launch period and the period after the fuel runs out. We will use suffices 1 and 2 for the distances travelled in these periods. The launch period: u = 0 a = 10.5 m/s 2 t = 6.6 s v =? Using v = 0 + 10.5 x 6.6 = 69.3 m/s Using = h 1 h 1 = 0 + 0.5 x 10.5 x6.6 2 = 228.7 m 3
After fuel runs out: a = g = -9.8 m/s 2 u = 69.3 m/s v = 0 s = h 2 Using 0 = 69.3 2 + 2 x (-9.8) x h 2 = 4802.49 19.6h 2 Therefore 245 m Therefore total height = h 1 + h 2 = 228.7 + 245 =473.7 m Tutorial Problems - From Applied Mechanics by Hannah & Hillier 1. A satellite-carrying rocket is launched vertically with constant acceleration 8 m/s 2 for the first stage lasting 100s. The acceleration increases to 18 m/s 2 for the second stage and remains constant for 60 s. The satellite then separates from the rocket and continues upwards freely under gravity (g = 9 m/s 2 ). Find its maximum altitude and total time taken to reach it. (317 km; 369 s) Angular Motion Angular (ω) Angular velocity is defined as the rate of change of angle per unit of time. Although rev/s is commonly used to measure angular velocity, in mechanics we use rad/s. Since one revolution is equal to 2 rads we can convert rev/s to rad/s by. Angular Acceleration (α) Angular acceleration is defined as the rate of change of angular velocity and has units of rad/s 2. Relationship Between Linear and Angular Motions In order to convert between these motions the following may be used: Equations of Motion The following relationships can be derived: It is also worth remembering that the angle turned by a wheel is the area under the angular velocity - time graph. Tutorial Problems 1. A wheel accelerates from rest to 3rad/s in 5 seconds. Sketch the graph and determine the angle rotated. (7.5 rads) 4
2. A wheel rotates from rest to 4rad/s in 4seconds. It then rotates at constant speed for 3 seconds and then decelerates to rest in 5 seconds. Sketch the velocity-time graph and determine: (a) the angle rotated, (b) the initial angular acceleration, (c) the average angular velocity. (30 rad, 1rad/s 2, 2.5rad/s) 5