Question

A ring of mass m can slide along a fixed rough vertical rod as shown. The ring is connected by a spring of constant K R where 2R is the natural length of spring. The other end of spring is fixed to the ground at point A at a horizontal distance of 2R from the base of the rod. 3R If the ring is released from a height of 2and it reaches the ground with a speed of 3gR The coefficient friction between the rod and ring is; 3R/2 2R A) 413-n2 ] B 413-2n2 )3-4n2] D) 2

Answer 1

Given mass of ring = m spring constant = > k = (4 mg/R) Natural length of the spring = l_0 = 2R speed of ring when reaching ground, v = squareroot 3g R Coefficient of fraction = mu At angle theta mg + F_s middot sin theta - f = m times a (where a is acceleration) and f = mu F_s middot cos theta mg + F_s middot sin theta - mu F_s middot cos theta = m times a also F_s = k times Delta x where Delta x = 2R(1/cos theta - 1) so mg + 2Rk (1/cos theta - 1) [sin theta - mu cos theta] = mx times dv/dt dg/dy mg + 2Rk (squareroot 4R^2 + y^2/2R - 1) (y/squareroot 4R^2 + y^2 - mu 2R/squareroot 4R^2 = y^2) = m middot d^2 y/dt^2 [mg = squareroot 4R^2 + y^2 - 2R) y - 2 mu R/squareroot 4R^2 + y^2) K] dy = mv dv