A 0.440-kg pendulum bob passes through the lowest part of its path at a speed of 3.38 m/s.(a) What is the magnitude of the tension in the pendulum cable at this point if the pendulum is 77.0 cm long? N(b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (Enter your answer to at least one decimal place.) The correct answer is not zero.°(c) What is the magnitude of the tension in the pendulum cable when the pendulum reaches its highest point? Your response differs from the correct answer by more than 100%. N
Question
A 0.440-kg pendulum bob passes through the lowest part of its path at a speed of 3.38 m/s.(a) What is the magnitude of the tension in the pendulum cable at this point if the pendulum is 77.0 cm long? N(b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (Enter your answer to at least one decimal place.) The correct answer is not zero.°(c) What is the magnitude of the tension in the pendulum cable when the pendulum reaches its highest point? Your response differs from the correct answer by more than 100%. N
Solution 1
(a) The tension in the pendulum cable at the lowest point of its path can be calculated using the formula for the centripetal force, which is F = mv^2/r, where m is the mass of the bob, v is the speed of the bob, and r is the length of the pendulum. The tension is the sum of the centripetal force and the weight of the bob (mg), so:
T = mv^2/r + mg
Substituting the given values:
T = (0.440 kg)(3.38 m/s)^2/(0.770 m) + (0.440 kg)(9.8 m/s^2) T = 6.22 N + 4.31 N T = 10.53 N
(b) At the highest point of its path, the pendulum bob is momentarily at rest, so the only forces acting on it are gravity (mg) and the tension in the cable (T). These forces are balanced, so the angle θ that the cable makes with the vertical can be calculated using the equation for the tangent of the angle, which is tan(θ) = opposite/adjacent. In this case, the opposite side is the horizontal component of the tension (Tsinθ = mg) and the adjacent side is the vertical component of the tension (Tcosθ = T). Solving for θ gives:
tan(θ) = mg/T θ = arctan(mg/T)
Substituting the given values:
θ = arctan((0.440 kg)(9.8 m/s^2)/10.53 N) θ = 23.1°
(c) At the highest point of its path, the pendulum bob is momentarily at rest, so the only force acting on it is gravity. Therefore, the tension in the cable is equal to the weight of the bob, which is:
T = mg T = (0.440 kg)(9.8 m/s^2) T = 4.31 N
Solution 2
(a) The tension in the pendulum cable at the lowest point of its path can be calculated using the formula for the centripetal force, which is the net force acting on the pendulum bob at this point. The centripetal force is given by the equation F = mv^2/r, where m is the mass of the bob, v is its speed, and r is the length of the pendulum. The tension T at the lowest point is the sum of the centripetal force and the weight of the bob (mg), so we have:
T = mv^2/r + mg
Substituting the given values:
T = (0.440 kg)(3.38 m/s)^2/(0.770
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