The point moves in a circle with a radius of R = 4 m. The law of its motion is expressed by the equation s = 8-2t^2.

The point moves in a circle with a radius of R = 4 m. The law of its motion is expressed by the equation s = 8-2t2. Determine the moment of time t when the normal acceleration of a point is equal to 9 m / s 2.

Data: R (radius of the circle) = 4 m; the law of motion of the point S (t) = 8 – 2t ^ 2; an (required normal acceleration) = 9 m / s2.

1) Required speed of movement: an = V ^ 2 / R, whence V = √ (R * an) = √ (4 * 9) = 6 m / s.

2) The law of change in the speed of a point: V (t) = S (t) ‘= (8 – 2t ^ 2)’ = 0 – 4t = -4t.

3) The sought moment in time: 6 = -4t, whence t = 6/4 = 1.5 s.

Answer: The normal acceleration of the set point will be 9 m / s2 after 1.5 s.