Question

PROBLEM 10 A point describes a circumference according to law s (t) = t³ + 2t² where s measured meters along the circumference and, in seconds. If a total acceleration of the point is 16 2 m / s2, when t = 2 s.calculate the radius R of the circumference?

Answer 1

`R = 25 " m"`

Explanation:

The acceleration vector (radial and tangential) is:

`mathbf a = - (dot s^2)/R mathbf \ mathbf hat r + ddot s \ mathbf hat theta`

Computing derivatives:

`s (t) = t^3 + 2t^2`

`dot s (2) = 3t^2 + 4t |_(t = 2) = 20`

`ddot s (2) = 6t + 4 |_(t = 2) = 16`

`mathbf a(2) = - 400/ r mathbf \ mathbf hat r + 16 \ mathbf hat theta`

CORRECTING TYPO IN QUESTION:

`| mathbf a(2)| = 16color(red)(sqrt 2) = sqrt ( (- 400/ R )^2 + (16)^2 )`

Solves as: `R = 25 " m"`