A ball weighing 1 kg rotates uniformly on a 0.6 m long thread with an angular velocity of 6 rad / sec

A ball weighing 1 kg rotates uniformly on a 0.6 m long thread with an angular velocity of 6 rad / sec (in the vertical plane) find the thread tension at the lower point of the trajectory.

To find the tension force of the thread T at the bottom point of the trajectory, on which a ball uniformly rotates in a vertical plane with a mass of m = 1 kg, we use Newton’s second law, according to which the resultant force is: m ∙ a = T – m ∙ g, where the proportionality coefficient g = 9.8 N / kg. Then T = m ∙ (a – g), where the centripetal acceleration is determined by the formula: a = a (c) = v² / R, v is the speed of the ball, R is the radius of the orbit of rotation equal to the length of the thread. From the condition of the problem it is known that the length of the thread is L = R = 0.6 m, the angular velocity ω = 6 rad / sec. We get T = m ∙ (ω² ∙ L – g), since v = ω ∙ R. Substitute the values ​​of physical quantities in the calculation formula and find the force T = 1 ∙ (6² ∙ 0.6 – 9.8); T = 11.8 N.

Answer: the thread tension at the bottom of the trajectory is 11.8 N.