At what speed is it necessary to rotate a body with a mass of m = 200 g on a thread with a length of l = 40 cm so that the thread does not break at the bottom point of the trajectory if it withstands the maximum tension force T = 6 N?
m = 200 g = 0.2 kg.
g = 9.8 m / s ^ 2.
l = 40 cm = 0.4 m.
T = 6 N.
Two forces act on the body at the lowest point during rotation: the gravity force m * g directed vertically downward, the tension force of the thread T directed vertically upward.
Let’s write Newton’s 2 law for a body at the lowest point of rotation: m * a = T – m * g.
Centripetal acceleration, and is determined by the formula: a = V ^ 2 / l, where V is the speed of rotation, l is the radius of rotation.
m * V ^ 2 / l = T – m * g.
V ^ 2 = (T – m * g) * l / m.
V = √ ((T – m * g) * l / m).
V = √ ((6 N – 0.2 kg * 9.8 m / s ^ 2) * 0.4 m) / 0.2 kg) = 2.85 m / s.
Answer: the rotation speed should not exceed V = 2.85 m / s.
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