CHAPTER 13 LINEAR AND ANGULAR MOTION
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1 CHAPTER 13 LINEAR AND ANGULAR MOTION EXERCISE 73, Page A pulley driving a belt has a diameter of 360 mm and is turning at 700/π revolutions per minute. Find the angular velocity of the pulley and the linear velocity of the belt assuming that no slip occurs. Angular velocity ω = n, where n is the speed of revolution in revolutions per second, i.e. n = π revolutions per second. Thus, angular velocity, ω = 700 = 90 rad/s 60π The linear velocity of a point on the rim, v r, where r is the radius of the wheel, i.e. r = 360 = 180 mm = 0.18 m Thus, linear velocity, v r = = 16. m/s. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm. Determine the angular velocity of the wheels of the bicycle and the linear velocity of a point on the rim of one of the wheels Linear velocity, v = 36 km/h = m/s = 10 m/s 3600 (Note that changing from km/h to m/s involves dividing by 3.6) Radius of wheel, r = 500 = 50 mm = 0.5 m Since, v r, then angular velocity, ω = v = 10 = 40 rad/s r
2 EXERCISE 74, Page A flywheel rotating with an angular velocity of 00 rad/s is uniformly accelerated at a rate of 5 rad/s for 15 s. Find the final angular velocity of the flywheel both in rad/s and revolutions per minute. Angular velocity, ω 1= 00 rad/s, angular acceleration, α = 5 rad/s and time, t = 15 s. Final angular velocity, ω 1+ αt = 00 + (5)(15) = = 75 rad/s In revolutions per minute, 75 rad/s = = 850 π π rev/min or 66 rev/min. A disc accelerates uniformly from 300 revolutions per minute to 600 revolutions per minute in 5 s. Determine its angular acceleration and the linear acceleration of a point on the rim of the disc, if the radius of the disc is 50 mm. Initial angular velocity, ω 1= and final angular velocity, ω = ω 1+ αt from which, 300 = 10πrad/s = 0πrad/s 60 ω ω1 0π 10π 10π angular acceleration, α = = = = 0.4π rad/s or 1.57 rad/s t 5 5 Linear acceleration, a = rα = (0.5)(0.4π) = 0.1π m/s or m/s 10
3 EXERCISE 75, Page A grinding wheel makes 300 revolutions when slowing down uniformly from 1000 rad/s to 400 rad/s. Find the time for this reduction in speed. ω 1+ω Angle turned through, θ= t hence 300 = t i.e. 600π = 700t from which, time, t = 600 π =.693 s 700. Find the angular retardation for the grinding wheel in question 1. ω 1+ αt from which, ω ω angular acceleration, α = = = = -.8 rad/s t i.e. angular retardation is.8 rad/s 3. A disc accelerates uniformly from 300 revolutions per minute to 600 revolutions per minute in 5 s. Calculate the number of revolutions the disc makes during this accelerating period. Angle turned through, 300 π 600 π ω 1+ω θ = t = (5) rad However, there are radians in 1 revolution, hence, number of revolutions = = ( 5) = revolutions 11
4 4. A pulley is accelerated uniformly from rest at a rate of 8 rad/s. After 0 s the acceleration stops and the pulley runs at constant speed for min, and then the pulley comes uniformly to rest after a further 40 s. Calculate: (a) the angular velocity after the period of acceleration, (b) the deceleration, (c) the total number of revolutions made by the pulley. (a) Angular velocity after acceleration period, ω 1+ αt = 0 + (8)(0) = 160 rad/s (b) ω 3 + αt from which, ω3 ω angular acceleration, α = = = - 4 rad/s t 40 i.e. angular deceleration is 4 rad/s ω 1+ω (c) Initial angle turned through, θ 1 = t = (0) = 1600 rad = 1600 rev At constant speed, angle turned through, θ = 160 rad/s ( 60)s = 1900 rad = 1900 rev Angle turned through during deceleration, θ 3 = (40) = 300 rad = 300 rev Hence, total number of revolutions made by the pulley = θ 1 + θ + θ 3 = = 4000 = 1000 π rev or 380 rev 1
5 EXERCISE 76, Page A car is moving along a straight horizontal road at 79. km/h and rain is falling vertically downwards at 6.4 km/h. Find the velocity of the rain relative to the driver of the car. The space diagram is shown in diagram (a). The velocity diagram is shown in diagram (b) and the velocity of the rain relative to the driver is given by vector rc where rc = re + ec rc = ( ) + = 83.5 km/h and θ= tan = (a) (b) i.e. the velocity of the rain relative to the driver is 83.5 km/h at 71.6 to the vertical.. Calculate the time needed to swim across a river 14 m wide when the swimmer can swim at km/h in still water and the river is flowing at 1 km/h. At what angle to the bank should the swimmer swim? The swimmer swims at km/h relative to the water, and as he swims the movement of the water carries him downstream. He must therefore aim against the flow of the water at an angle θ shown in the triangle of velocities shown below where v is the swimmers true speed. v = 1 = 3km/h = m/min = 8.87 m/min 60 13
6 Hence, if the width of the river is 14 m, the swimmer will take = minutes = 4 min 55 s In the above diagram, sin θ = 1 from which, θ = 30 Hence, the swimmer needs to swim at an angle of 60 to the bank (shown as angle α in the diagram. 3. A ship is heading in a direction N 60 E at a speed which in still water would be 0 km/h. It is carried off course by a current of 8 km/h in a direction of E 50 S. Calculate the ship s actual speed and direction. In the triangle of velocities shown below (triangle 0AB), 0A represents the velocity of the ship in still water, AB represents the velocity of the water relative to the earth, and 0B is the velocity of the ship relative to the earth. Total horizontal component of v = 0 cos cos 310 =.46 Total vertical component of v = 0 sin sin 310 = 3.87 Hence, v = ( ) + =.79 km/h, and θ= tan = Hence, the ships actual speed is.79 km/h in a direction E 9.78 N 14
7 EXERCISE 77, Page 178 Answers found from within the text of the chapter, pages 171 to 177. EXERCISE 78, Page (b). (c) 3. (a) 4. (c) 5. (a) 6. (d) 7. (c) 8. (b) 9. (d) 10. (c) 11. (b) 1. (d) 13. (a) 15
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