WJEC Physics for A2: Student Bk

14 We can use this result and lim sin θ θ θ → 0 = 1 to calculate the instantaneous acceleration at the top of the arc, i.e. when θ = 0 . The answer is the very simple equation a = v 2 r which, applying v = r ω we can write as a = r ω 2 Because this acceleration is directed towards the centre of the circle it is referred to as the centripetal acceleration (from the Latin, to seek the centre). Example The international space station (ISS) orbits at a mean height, h , of 412 km with an orbital period, T , of 92.69 minutes. Calculate its acceleration. [Mean radius of the Earth, R E = 6371 km .] Answer The angular velocity of the ISS = 2 π T ∴ a = r 2 π = 4 π 2 T T 2 2 ( R E + h ) = 4 π 2 (6371 + 412) × 10 3 m (96.29 × 60 s) 2 = 8.02 m s −2 . (See Study point) 3.1.4 Centripetal force We can now use Newton’s laws to move from kinematics to dynamics; we can explore what makes a body move in a circle at a steady speed. N1 and N2 tell us that, if an object is accelerating, the resultant of all the forces on it cannot be zero; there must be a resultant force in the direction of the acceleration, i.e. towards the centre of the circular path. We can calculate the magnitude of this centripetal (resultant) force using the N2 equation, F = ma . From above: The centripetal force, F = mv 2 r or F = mr ω 2 . Some examples of centripetal forces 1. Every one of the HI regions (the pink clouds in NGC1566 – the spiral galaxy in Fig. 3.1.10) is acted on by the gravitational fields of all the other objects in the galaxy. The resultant force on each HI region (as well as on the stars) is towards the centre of the galaxy. 2. The resultant of the tension and gravitational forces on the riders on the swing carousel is towards the axis of the carousel. 3. The sideways grip of a car’s tyres on a horizontal road as it negotiates a bend (Fig. 3.1.11) is towards the centre of the bend. It is worth examining that last one. Tyres have a maximum grip, F max , which is normally expressed as a fraction, μ , of the normal contact force, C , between the car and the road. On a horizontal road, C is equal (and opposite) to the weight, mg , of the car. i.e. F max = μ C Typically μ ~ 1 on a dry road and 0.2 on a wet road. Using this information we can estimate the maximum safe driving speed around a bend. Study point Check that the formula 〈 a 〉 = v 2 sin θ r θ gives the same answers as in the section above. 3.1.6 Self-test Use the formula a = v 2 r to calculate the acceleration at point Y in Fig. 3.1.7. Study point It is no accident that the ISS acceleration is just a little smaller than g at the surface of the Earth. This is because of the inverse square law of gravitation and will be taken up in Section 4.2. Fig. 3.1.10 M51 – the Whirlpool galaxy F v Fig. 3.1.11 Car on a bend WJEC Physics for A2 Level: Unit 3