WJEC Physics for A2: Study and Rev Guide

A ball, of mass 0.100kg , is attached to a string and whirled in a horizontal circle of radius 0.30m at a rate of 1.1 revolutions per second. The string sweeps out a cone as in Fig. 3.1.5 (a). Calculate the angle of the string to the vertical and the tension, S . Answer The resultant force on the bob provides its centripetal acceleration, So F res = mr w 2 = 0.100kg × 0.30m × (2 p × 1.1 s −1 ) 2 = 1.43N. This must be the resultant of the pull, S , of the string and the pull of gravity, mg , on the bob. See Fig. 3.1.5 (b) and (c). From (c) we deduce that S = F 2 res + ( mg ) 2 = 1.43 2 + 0.981 2 N = 1.73 N and q = tan −1 mr w 2 mg = tan −1 1.43 0.981 = 56° G In a ‘conical pendulum’ set-up (see example) the string is 0.60m long and is always at 30° to the vertical. The bob’s mass is 0.100kg . Calculate: a) the tension in the string (consider vertical force components) b) the radius of the bob’s circular path c) the bob’s speed. QUICKFIR QUICKFIRE QUICKFIRE 1. Fig. 3.1.6 shows a simple ‘chain drive’ on a bicycle. chain ring (toothed wheel) back sprocket (toothed wheel) axle (turns back wheel) axle (turned by pedals) chain 8.09cm 3.23 cm Fig. 3.1.6 Bicycle chain drive A cyclist pedals so as to turn the chain ring at a rate of 1.50 revolutions per second. Calculate: (a) the angular velocity of the chain ring (b) the speed at which the links in the chain are made to move (relative to the bicycle) (c) the angular velocity of the back sprocket (and back wheel) (d) the speed at which the bicycle is travelling, if the diameter of the back wheel (to the tread of the tyre) is 0.67m . UICKFIRE 11 3.1 Circular motion