Tiêu đề 06 Circular Motion Tutorial Solutions.pdf ✅

St. Andrew’s Junior College H2 Physics 6-3 Diagram FBD Force Equations that may be used to solve such problems Object travelling in a loop-the- loop N mg mg + N = m v 2 r Derive the minimum velocity, v, such that the object can complete the circular motion. When N = 0, mg = m v 2 r v = √rg A mass tied to a cord under non-uniform vertical circular motion: At the top of the circular motion: T mg mg + T = m v 2 r Derive the minimum velocity, v, such that the object can complete the circular motion. When T = 0, mg = m v 2 r v = √rg At the bottom of the circular motion: T mg Force Equations T - mg = m v 2 r Using the COE, derive an expression to show the mathematical relationship between vbottom and vtop By COE, KEbottom = KEtop + GPEtop 1⁄2 mvbottom2 = 1⁄2 mvtop2 + mg(2r) 1⁄2 vbottom2 = 1⁄2 vtop2 + 2gr Conclusion: vbottom > vtop Show your mathematical reasoning that it is more likely for the string to break at the bottom than at the top of the non-uniform circular motion. Bottom Top Tbottom - mg = (m v 2 r )bottom Tbottom = (m v 2 r )bottom + mg Ttop + mg = (m v 2 r )top Ttop = (m v 2 r )top - mg Conclusion: Comparing the 2 equations and knowing vbottom > vtop, it can be concluded that Tbottom > Ttop. r v vtop r vbottom r
St. Andrew’s Junior College H2 Physics 6-4 TUTORIAL 6: MOTION IN A CIRCLE SOLUTIONS Level 1 Solutions 1(a) 300 rpm = 5rps = 5 x 2π s-1 = 10π s-1 = 31.42 s-1 = 31.4 rad s -1 [1] (b) Since a = rω2 , and ω is constant, a α r. Therefore, it is a linear graph. [1] (c) v1 = r1ω = (0.030)(31.42) = 0.943 m s-1 v2 = r2ω = (0.050)(31.42) = 1.57 m s-1 [1] [1] 2 {Tension [1], Weight [1] } 3(a) r = 0.2 2 = 0.1 m ω = 600 rpm = 10 rps = 62.831 rad s-1 Fc = mrω2 = 10-4 (0.1) (62.831)2 = 3.9478 x 10-2 N = 3.95 x 10-2 N (3.s.f) [1] (b)(i) Fmax = mrωmax 2 0.1 = 10-4 (0.1) ω2 ω2 = 104 ω = 100 rad s-1 [1] [1] (ii) v = rω = 0.1 (100) = 10 m s-1 At the moment the pea leaves the wheel, its linear velocity is the same as the linear velocity of the wheel. [1] a / ms-2 r / m Gradient = ω 2 Tension a = v 2 r Weight Note: A common mistake is including the centripetal force. Students must remember that centripetal force is a resultant force that does not exist by itself but is the resultant of the forces acting on the object. In this case, the horizontal component of the tension.