HCI 06 Circular Motion Solutions (Discussion Questions)

Circular Motion Discussion Questions Suggested Solutions D1 Given mass, m = 2.0 × 10-3 kg (S.I unit) and time for 3 revolutions, 3T =3.14 s  T =3.14/3 s (a)  = t = 2𝜋 3.14/3 × (2) = 12.0 rad = 12.0 − 2𝜋 = 5.7 rad (b) r = 0.05 m 𝑣 = 𝑟𝜔 = 2𝜋 3.14/3 × 0.05 = 0.30 𝑚𝑠−1 (c) or = 1.800 = 1.8 m s-2 (d) Note: Frictional force, f, provides the centripetal force, Fc. f = Fc = mac = (2.0 × 10-3)(1.800) = 3.600 × 10-3 = 3.6 × 10-3 N D2 Tension, N0820040 ..xkT === Tension provides the centripetal force, r vmT 2 = -18×0.70= = =10.58 =11 m s0.050 Trv m Answer: A D3 The external forces acting on the pendulum are tension (strictly speaking, force of string on pendulum) and weight. The horizontal component of the tension provides the centripetal force for the pendulum to turn right. (Note that the centripetal force should not be drawn as a third force.) Answer: C D4 Frictional force of road on car F provides for centripetal force for car to round the corner. r vmF 2 = Since m and r remain the same, F  v2 2)v v(F F dry w et dry w et = 2 202 1 )v( w et= 2 20=w etv Answer: D 2ra= r va 2 = 2 2 100.5 )3001.0( −= Writing this statement is necessary to justify why friction is equated to the centripetal force in the next step of working.

D5 (a) (b) (c) Diagram 1. Frictional force of wall on person keeps him/her from sliding down. 223.00 5.00 75.0 ms −=  =2 ca = r ω 60 75 4500 N=  =cF = ma Normal contact force of wall on person provides for the centripetal acceleration. D6 (a) (b) For m1, T1 – T2 = m1a1 (Newton’s second law) 2 1 21 1 sm64052 9254 −=−=−= .. .. m TTa For m2, T2 = m2a2 (Newton’s second law) 2 2 2 2 sm83053 92 −=== .. . m Ta r va 2 = 1 111 sm80001640 −=== ...rav 1 222 sm0104131830 −==== ....rav D7 (a) (b) (c)(i) (ii) (iii) The tension in the string created by the weight, W of the washers provides the centripetal force for the bung to perform circular motion. If we let M = mass of washers and m = mass of rubber bung, then = = = 22()Tension W Mg mr T 220.035 9.81 (0.57)( ) 12.618.2 20 mg  = = If glass rod is twirled at just the right pace the paper clip can be maintained in a position just below the bottom of the glass tube. This ensures that the radius is kept constant at a known radius. 3 1 3 95 10 9.81 0.57 6.49 m s12.6 10 Mgrv m − − −   = = =  There must exist an upward vertical component of the tension to balance the weight of the rubber bung. If the string was purely horizontal no such component could exist. In the vertical direction, forces acting on rubber bung sin mgT  = Hence,  − − ==  3 3 12.6 10 95 10 mgsin Mg θ = 7.62 ⁰ T2 T1 m1 T2 m2

D8 (a) (b) String is just taut when particle reaches C → Tension T = 0 Weight of particl