a) The radius of the circle traced is r = l sin(θ), where l is the string length. Substituting l = 3 m and θ = 20°, we have sin(θ) ≈ 0.34, and r ≈ 0.34 × 3 m ≈ 1 m.
Answer: r ≈ 1 m.
b) The string tension T should ensure the uniform circular motion of the pendulum. It is most easily found by balancing the vertical forces acting on the pendulum, since the pendulum does not move in the vertical direction. The vertical upward projection of the string tension is T cos(θ), and it should balance the vertical downward force of gravity, which is equal mg, where m = 3.1 kg is the pendulum mass, and g ≈ 9.8 m/s2 is the acceleration of gravity. Hence, T cos(θ) = mg, whence T = mg/cos(θ). We have cos(θ) = cos(20°) ≈ 0.94. Substituting the values of other quantities, we get T ≈ 3.1 × 9.8/0.94 kg m/s2 ≈ 0.32 kg m/s2 = 0.32 N.
Answer: T ≈ 0.32 N.
c) The period of the motion t of a conical pendulum is found by equating the centripetal acceleration of the pendulum and the horizontal projection of the tension T – the only force acting in the horizontal direction. We have T sin(θ) = mv2/r, where v is the circular velocity. Hence, v=√(rTsin ϑ⁄m). The period is given by t=2π r⁄v=2π√(r m⁄T sinϑ). Expressing r in meters, m in kilograms, and T in Newtons, we obtain the period in seconds. Substituting the numbers found previously, we obtain t≈2π√((1×3.1)⁄((0.32×0.34) )) s≈33.5s.
Answer: t ≈ 33.5 s.