How to Calculate Uniform Circular motion in horizontal xy plane?

Question

A particle moves horizontally in uniform circular motion, over a
horizontal xy plane. At one instant, it moves through the point at coordinates (4.00 m,
4.00 m) with a velocity of iˆ 00 . 5 − m/s and an acceleration of jˆ +12.5 m/s2. What are the
(a) x and (b) y coordinates of the centre of the circular path?

Answer 1

If I understand your conditions correctly, in (x,y) notation, v = (0.5,0) and a = (0,12.5). Centripetal acceleration a is toward the center. Since a is along the y axis, the center is directly above the point (4,4). And since a = v^2/r, r = v^2/a = 0.02 m. Thus the center is at (4 m, 4.02 m). Of course, with those spaces and the floating minus sign or dash, I suppose "iˆ 00 . 5 − " could easily be something other than 0.5. If so, just substitute it for v to get the correct value for r.

Answer 2

the pressure is the same as mg and this equals M v^2/R for the cork. yet you've said that the string sags somewhat once you rotate it slowly. There are 2 outcomes going one. element of the pressure (the vertical aspect, T cos theta ) ought to help the burden of the cork Mg and section ( the horizontal section, T sin theta) ought to make it pass in a circle (Mv^2/r) in case you've performed aspects of forces you should take care of to appreciate, if no longer, that is tremendous.