Question

Consider a damped pendulum consisting of a ball of mass M and radius a suspended by a fine wire of length L below a pivot. The ball is moving through a fluid of viscosity y, and this retards its motion. It is assumed that the ball moves sufficiently slowly that Stokes law applies. Under these conditions, the equation describing the motion of the sphere is given by ML =-Mg sin 0 - 6xual, The initial conditions are 012-= 0, and p=0. dt0 (a) Choose an appropriate reference time scale and render the equation and initial conditions dimensionless. Show that the dimensionless solution depends only on two dimensionless parameters. What are the physical meanings of these parameters? (b) You are assigned the problem of adjusting the fluid viscosity so as to bring the sphere to rest as rapidly as possible. Recognizing that if u= 0, the pendulum will oscillate forever, and if u is very large, the sphere will take a long time to get to the equilibrium position, estimate the correct value of the viscosity for critical damping via scaling. Estimate the characteristic time required to reach the equilibrium value. (c) For small values of the initial amplitude 0. , sin Oz0. In this limit, solve the differential equation and compare the exact solution to that estimated by the dimensional analysis above.

Answer 1

Solution 6 'The Equation of motion you have : M2 d’O -- Mg sinco) dt o --GAM" L doo - o For small oscillation we can gets zinco – hef us rewrite squalion : mig GA - ML 28=6aua M & W2 = 9/2 ľ And define z = not, such that t'i 6 dimension-les. e de ce world cup

. We - we do Qub, all yn Egn :- we ont as we do not w*9=0 => det et do +0=0 + 25 d9 +0 -0 — where ß=rw 8=37M/M w=ugh t' = ut Now we write solution of eqn @ which ů now in dimension-len forma writting Auxillary egna 02 +2 Br +1=0

4= -25.4 $1492y = =P+ PP1 het e = UBE then solna O ct') = A & Bt)Z! 7 Be-B-2)21 OCH) = ēBz! [A&t='+ Bent'] Applying initial conditions of z=0 (ie zl=0) O=0. por A+8 -0. —© of z'o, dol=0 dit' A(2-3) - B (+2+3) =0 @ A=B (2 +B) [m-B) Elit + B = 0. u s ß

B = 0o (2-B) ar in Aa O. (-12+B) 22 tere Final Solutions B7) OU') = 0,2.82"[map] 2 * 462-f3ē *u] OC)= 0 EFt (2.2 Gsh(*+'+ap sinbeatus] Oce")= 0 eft' [n & h (=4+Bainh (t!)] dimension, les group Final solution : (Realy we have two in the solution). The End