WJEC Physics for A2: Student Bk

13 3.1 Circular motion 3.1.3 Centripetal acceleration Consider the object in Fig. 3.1.7. It is moving at a constant speed of 20 m s −1 with in a circle of radius 5 m . We’ll calculate the mean acceleration between points A and B , i.e. π 4 either side of the vertical line OY . The horizontal component of the velocity is unchanged between A and B at 20 cos π 4 m s − 1 to the left. The vertical component of velocity at A = 20sin π 4 m s − 1 in direction OY. The vertical component of velocity at B = 20sin π 4 m s − 1 in direction YO. = –20sin π 4 m s − 1 in direction OY ∴ Δ v = v B – v A = –20sin π 4 – –20sin π 4 = –40sin π 4 m s − 1 in direction OY . To calculate the mean acceleration we need to know the time interval between A and B . From the result of Self-test 3.1.4, we can now write: So the mean acceleration is given by: 〈 a 〉 = Δ v = – 40sin π 4 Δ t π 8 = −72 m s − 1 (2 s.f.) in the direction OY You should check this result. The figure of 72 m s −1 is not so important but the direction is: the mean acceleration is towards the centre of the circle. If you do Self-test 3.1.5, you will find a result with a very similar magnitude (slightly greater) but in the same direction: there is always a mean acceleration around the point Y in the direction YO. We’ll now build on this result and calculate the instantaneous acceleration at point Y itself. To do this we’ll need to use the result lim sin θ θ θ → 0 = 1 , which follows from the definitions of θ (in rad ) and sin θ . Fig. 3.1.8 is just a redrawn Fig. 3.1.7 with general values of the variables. We could use components again but, for flexibility, we’ll use the vector diagram, Fig. 3.1.9, to calculate Δ v . The magnitudes of v A and v B are equal and as they make the same angles to the horizontal. ∴ by symmetry, v B − v A is vertical and Δ v = v B − v A = 2 v sin θ The arc length AB is given by: arc AB = 2 r θ ∴ The time Δ t , between A and B is given by: Δ t = Arc AB = 2 r θ v v Hence the mean acceleration between A and B is given by: 〈 a 〉 = Δ v = 2 v sin θ × v = v 2 × sin θ Δ t 2 r θ r θ and the direction is downwards in Fig. 3.1.8, i.e. always towards the centre of the circle. 20 m s − 1 20 m s − 1 A B 5 m 44 O Y A B O v v r v A v B v B − v A Fig. 3.1.7 What’s the acceleration? 3.1.4 Self-test Use time = distance speed to show that the time interval between A and B in Fig. 3.1.7 is π 8 s. 3.1.5 Self-test Repeat the calculation of the mean acceleration for two points which are π 6 on either side of the vertical. MATHS TIP Use your calculator to determine the ratio sin θ θ , for θ (in rad) = 1.0, 0.1, 0.01, 0.001 and 0.0001 and show that the ratio approaches 1. Fig. 3.1.8 Centripetal acceleration Fig. 3.1.9 v B − v A