1. Velocity and Acceleration

Transcription

1 1. Velocity and Acceleration [This material relates predominantly to modules ELP034, ELP035] 1.1 Linear Motion 1. Angular Motion 1.3 Relationship between linear and angular motion 1.4 Uniform circular motion - (acceleration) 1.1 Linear Motion Linear displacement and distance The linear displacement is the length moed in a gien direction - it is a ector quantity. The magnitude of the displacement is the distance - a scalar quantity. Linear elocity and speed The linear elocity is the rate of change of displacement with time. As displacement is a ector so elocity is a ector. The magnitude of the elocity is speed. It is the rate of change of distance with time - hence it is a scalar. If a body moes with uniform elocity then it must moe in a fixed direction with constant speed. The aerage speed of a body is the total distance moed diide by the total time taken. Distance - time cure A graph plotted for distance (s) against time (t), might look like that in Figure 1.1:

2 distance s A B C Figure 1.1: Distance-Time Cure time t As speed is rate of change of distance with time, the slope, gradient, of the s/t cure is the speed. Oer the linear section OA of the cure the speed must be uniform. Between A and B the gradient is becoming less and less, hence the body is slowing down. At B the body is stopped (distance is not increasing) and remains at rest between B and C. Linear acceleration The linear acceleration of a body is the rate of change of linear elocity with time. It is a ector. If acceleration is uniform the speed must be increasing by equal amount in equal time interals. Worked example 1.1 A car is traelling along a straight road at 13 m/s. It accelerates uniformly for 15 s until it is moing at 5 m/s Solution change in elocity d Acceleration = = time taken 5 13 = 15 = 0.8m Speed - Time cure

3 A graph of speed () of a body plotted against time (t) might be as shown by the graph in figure 1.: D speed C B A t1 Figure 1.: Speed-Time Cure t time t As acceleration is rate of change of speed () with time (t), the slope, gradient, of the /t cure is the speed. In Figure 1. the gradient between A and B is increasing - hence acceleration - is increasing, between B and C it is constant, between C and D it is decreasing. If in the small time interal, the speed is. The distance coered in the time is ds = = The total distance s traelled in the time interal between t 1 and t is the integral of this i.e. s = ds = ds This integral is the same as the area under the cure. Thus the distance traelled in any time interal is the area under the /t cure between the start and end time. t t 1 Equations for linear uniformly accelerated motion If a body that is moing in a straight line and started with initial speed u undergoes a uniform acceleration a for a time t until its elocity is, then the speed time cure would look like that in Figure 1.3: 3

4 speed u Figure 1.3: Uniformly accelerated linear motion t time t Since a is uniform, its magnitude is change in speed a = change in time at = u u t = a = u + at = d u = t Equation 1.1 In this case, the aerage speed will be the speed at t/. Hence: u + aerage speed = Since the distance done = aerage speed time then s = ( u + ) t Equation 1. Substituting Equation 1.1 for into Equation 1. gies ( u + u + at) s = 1 s = u + at t Equation 1.3 Substituting Equation 1.1 for t into Equation 1. gies 4

5 s = as = ( u + )( u) + u = as u a Equation 1.4 These four equations are the equations for linear, uniformly accelerated motion. They all contain 4 unknowns, you must know three before you can find the fourth. Worked Example 1. A car starts from rest and accelerates in a straight line at 1.6 m/s for 10s. a) What is its final speed? b) How far has it traelled in this time? If the brakes are then applied and it traels a further 0m before stopping c) What is the deceleration (retardation)? Solution a) Initial speed u = 0. Acceleration a = 1.6 m/s Time t = 10s Use Equation 1.1 to gie = u + at = = 16m b) Use Equation 1.3 to gie distance gone c) 1 s = ut + at 1 s = s = 80m 5

6 Initial speed u = 16 m/s Final speed = 0 Distance s = 0 m Use Equation 1.4 to find acceleration = u 0 = 16 + as + a 0 a = 6.4 m Deceleration is negatie acceleration so Deceleration = 6.4 m/s Worked Example 1.3 A hoist starts at ground leel and accelerates as 1. m/s for 5s. It then moes with uniform speed for 10s and is finally brought to rest at the top of a building with a retardation (deceleration) of.0 m/s. a) Draw the speed time graph of the motion b) What is the height of the building? Solution a) We can draw the speed - time cure shape as in Figure 1.4 final speed (m/s) tfinal time (s) Figure 1.4: Speed-Time cure shape for Worked Example 1.3 6

7 But to complete this we need to find the unknowns, the final speed u and the time of the retardation. Initial speed u = 0.0 Acceleration a = 1. m/s Time t = 5 s Use Equation 1.1 to gie final speed = u + at = = 6.0m For the retardation time: Initial speed u = 6.0 m/s Final speed = 0 Acceleration a = -.0 m/s (NOTE: Negatie sign as retardation) Using Equation 1.1 to gie time = u + at t = ( u) a 6.0 = = 3s.0 We can now complete the speed time cure by putting in the alues, as shown in Figure speed (m/s) A B O C 18.0 time (s) Figure 1.5: Completed Speed-Time cure for Worked Example 1.3 b) The total distance traelled is the area under the speed-time cure. i.e. the area of the trapezium OABC 7

8 5 6 s = s = 84m + ( 10 6) Thus the height of the building is 84m Free fall under graity When a body falls to earth freely (without any other forces inoled) the acceleration is called acceleration due to graity. It is often gien the symbol g. Proided that air resistance is negligible all bodies, heay or light fall at the same acceleration. Although g aries ery slightly at different points on he earth g = 9.81 m/s can be used in most calculations, and will be used in the rest of these notes. (To four decimal places g = m/s.) Worked Example 1.4 A worker drops a hammer from the top of a 60m high building. If the speed of sound in air is 340 m/s, how long does the worker hae to shout down to warn colleagues (if his warning is to reach them before the hammer!) Neglect air resistance. Solution: The solution is, the length of time for the hammer to reach the ground minus the length of time it takes for the shouted warning to reach the workers on the ground. For the hammer: Initial speed u = 0 Acceleration a = 9.81 m/s Distance s = 60m Use Equation 1.3 to gie time. For the shout 1 s = ut + at 1 60 = t t = s t t = 1.34 = 3.5s = = 0.18s 8

9 Difference in trael time between hammer and shout: = 3.3 s The warning must be shouted within 3.3 seconds of dropping the hammer. 1. Angular Motion Angular speed and elocity Consider a point P moing along a line QP as shown in Figure 1.6 P θ 0 X Q Figure 1.6: Angular motion Let OX be a fixed line and θ be the angle made with OP at some time t. The angular elocity ω of P about O is the rate of change of θ with time in the sense of increasing θ. Hence dθ ω = Equation 1.5 dθ When the direction of increase of θ is not included, is called angular speed. The SI unit for angular speed is radians per second, rad/s. Angular acceleration If the angular elocity of the point P in Figure 1.6 is changing with time, then the angular acceleration, α, of P is the rate of change of its angular elocity: 9

10 In the sense of increasing θ. dω α = Equation 1.6 In SI units angular acceleration is measured in rad/s. If the angular acceleration is uniform, then when angular speed changes from ω 1 to ω in time t, its magnitude is ω ω α = 1 t 1.3 Relationship between linear and angular motion Equation 1.7 Relationship between linear speed and angular speed If a point P moe round a circle of radius r with constant linear speed,, (see Figure 1.7) then the angular speed, ω, will be constant at θ ω = t Equation 1.8 P θ Q 0 X r Figure 1.7: Circular Motion Where t is the time to moe from Q to P along the arc QP of the cure. Howeer, arc length QP is rθ when θ is measured in radians. Hence linear speed is length of arc QP rθ = = t t Equation 1.9 Substituting Equation 1.8 into Equation 1.9 leads to the this relationship for circular motion: 10

11 = rω linear speed = radius angular speed Equation 1.10 Relationship between angular speed and frequency of rotation Let P in Figure 1.7 rotate with constant frequency of n re/s. Since for each reolution turned the angle is π rad, then the number of radians turned through per second is πn. This is the angular speed. So we can write ω = πn angular speed = π reolutions per second Equation 1.11 Relationship between linear acceleration and angular acceleration By Equation 1.6 and Equation 1.10 and dω α = = rω α = d r as r is constant this can be written 1 d α = r d and as is linear acceleration a, a α = r a = rα Equation Uniform circular motion - (acceleration) Consider again a point X moing round a circle, radius r, at uniform speed. Since the direction of motion is changing form instant to instant, the elocity is changing and hence point X has an acceleration. 11

12 Let AB represent the elocity ector when X is at point Q and AC represent the elocity ector when X is at point P, as shown in Figure 1.8. The elocity ectors at points Q and P are tangential to the circle of motion, sides AB and AC are parallel to these ectors. B r P C θ/ θ/ 0 θ r Q X θ A Figure 1.8: Angular Acceleration The line BC represents the change in elocity as the point moes from Q to P. The aerage acceleration of P is length of BC a = time to trael from Q to P So length of arc QP = rθ time to trael from Q to P = rθ / Also length of BC = sin ( θ / ) a = r ( θ / ) sin θ / To find the acceleration of P at Q, let the angle θ/ tend to zero, so sin( θ / ) θ / and the acceleration is gien by 1

13 a = r Equation 1.13 When θ / 0, the direction of BC approaches the direction of QO, that is, towards the centre of the circle. Hence, the acceleration of a point moing round a circle with radius r, at constant speed is /r towards the centre of the circle. Worked Example 1.5 A flywheel, diameter 1.1m, rotating at 100 re/min slows down at a constant rate to 900 re/min in 30 s. Find: a) The initial angular speed b) The final angular speed c) The angular acceleration d) The initial speed of a point on the rim of the flywheel. Solution a) From Equation 1.11 angular speed ω = πn Initial n = 100 re/min = 100 /60 = 0 re ω = π 0 = 15.7rad b) Final n = 900 re/min = 900/60 = 15 re/s ω = π 15 = 94.rad c) Use Equation 1.7 ω ω α = = t 30 α = 1.05rad d) Equation 1.10 gies the linear speed, so initial speed on edge of flywheel is = rω 1.1 = 15.7 = 69.1m 13

14 Worked Example 1.6 The spin dryer in a washing machine is a cylinder with diameter 500mm. It spins at 900 re/min. Find a) the speed of a point on the edge of the drum b) the acceleration of a point on the edge of the drum Solution a) Frequency of rotation in re/s n = 900/60 = 15 re/s From Equation 1.11 ω = πn = π 15 = 94.re From Equation 1.10 the linear speed is = rω = 94. = 3.6m b) From Equation a = = 30m r / = ( ) Note how large this is, it is 7 times the acceleration due to graity! 14