Y13 Circular Motion

(a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                           2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown             2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

          Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                              5

  (b)         (i)            k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)         

the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                        3

MAXIMUM 3