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 particle provides for centripetal force L vmgm C 2 = LgvC = (shown) By conservation of energy, Initial total energy at the bottom = Final total energy at the top ( )Lgmvmvm C 22 1 2 1 22 += ( )LggLv 22 1 2 1 2 += gLgLv 42 += gLv 5= D9 (a) Let v be the tangential velocity at A and m be the mass of the roller coaster. Using conservation of energy, Initial KE + initial GPE = final KE + final GPE 1 2 𝑚𝑣𝑜 2 + 𝑚𝑔ℎ = 𝑚𝑣2 2 + 2 3 𝑚𝑔ℎ 1 2 𝑚𝑣𝑜 2 + 1 3 𝑚𝑔ℎ = 𝑚𝑣2 2 − − − −(1) The resultant force on the roller coaster at A provides the centripetal force, Using N2L, W – N = 𝑚𝑣2 𝑅 When N = 0, v is maximum 𝑚𝑔 = 𝑚𝑣2 𝑅 1 2 𝑅𝑚𝑔 = 𝑚𝑣2 2 − − − −(2) Sub (2) into (1), when v is maximum, vo will also be maximum: 1 2 𝑚𝑣𝑜 2 + 1 3 𝑚𝑔ℎ = 1 2 𝑅𝑚𝑔 𝑣𝑜 2 = 𝑅𝑔 − 2 3 𝑔ℎ 𝑣𝑜 = √𝑔(𝑅 − 2 3 ℎ) mg

(b) By conservation of energy, Initial total energy = Final total energy at B 1 2 𝑚𝑣𝑜 2 + 𝑚𝑔ℎ = 𝑚𝑔ℎ′ (just makes it to B  final KE = 0) 𝑔ℎ′ = 1 2 𝑣𝑜 2 + 𝑔ℎ ℎ′ = 1 2 𝑣𝑜 2 𝑔 + ℎ ℎ′ = 1 2 𝑅𝑔 − 2 3 𝑔ℎ 𝑔 + ℎ ℎ′ = 2 3 ℎ + 1 2 𝑅 D9 (c) The acceleration is not uniform. D10 ( ) =− =+ −= −−= -1 A C C 22 AC 22 At point C, particle's velocity is entirely horizontal. i.e. 3.0 m s By conservation of energy, Total energy at A = total energy at C KE GPE KE 1 (2 )2 5.0 ( 3.0) 4 v m v v mg R R − = = = = 22 2 c 0.408 m 3 22.1 m s0.408 g va R Answer: D D11 (a) (b) (c) (d) 1sm202020 07π2π2 −=== .. . T rv 22 50 2.2 34.6 N7.0 c mvF r = = = Considering the forces acting on Sasha at the top of the ride and using N2L: W – N = 2mv r For Sasha to feel weightless at the top of the ride, N = 0 gmr vm = 2 1sm29881907 −=== ...grv At the bottom of the Ferris wheel, 2 =−mv N mgr 2 2 2 50 9.81 981 NmvN mg m g r= + = =   = Note that ac = r2 is not applicable in this context. This is because the angular speed of the particle is not constant throughout its vertical circular motion. The centripetal acceleration varies as the speed of the particle changes. N W O

D12 (a) (b) (c) (d) t  =  6037 π2 = = 2.8303 x 10-3 = 2.83 x 10-3 rad s-1 v = r  = ).( 310830322 150 −     = 0.21227 = 0.212 m s-1  =t 31083032 32 π22 −      = . = 138.75 = 139 s or 232 6037 =t = 139 s