A very light rigid rod with a length of 0.546 m extends straight out from one en
ID: 2212231 • Letter: A
Question
A very light rigid rod with a length of 0.546 m extends straight out from one end of a meter stick. The combination is suspended from a pivot at the upper end of the rod as shown in the following figure. The combination is then pulled out by a small angle and released. (a) Determine the period of oscillation of the system. (b) By what percentage does the period differ from the period of a simple pendulum 1.046 m long? I tried using the formula T=2pi(L/g)^1/2 and got 2.5 seconds but it was wrong and I dont know what other formulas I can use. Please help!Explanation / Answer
If a thin rod is rotating at angular velocity w, the kinetic energy of a bit of mass dm (= lambda*dx) at distance x from the axis is given by: lambda*dx *v^2/2 where v = w*x Therefore, d(KE) = lambda*dx*(w^2)*(x^2)/2 So integrating over x from 0 to L gives: integral d(KE) = (w^2/2) * lambda * integral(x=0,L) [x^2]dx = (w^2/2) * lambda * L*3/3 So: KE = (w^2/2) * lambda * L^3/3 = (w^2/2) * (lambda * L) * L^2/3 = (w^2/2) * Mass * L^2/3 a) The potential energy PE is given by the potential energy of the center of mass of the rod: U = - MgL*cos(angle_from_vertical)/2 = - (MgL/2)*cos(a) KE = (w^2/2) * M * L^2/3 = M*L^2*w^2/6 E = KE + PE = M*L^2 * w^2 /6 - Mg*L*cos(a)/2 = (M*L^2/2) * (w^2/3 - (g/L)*cos(a)) = (M*L^2/6) * (w^2 - (3g/L)*cos(a)) Differentiating wrt time: 0 = 2w*dw/dt + (3g/L)*sin(a)*da/dt = 2w*dw/dt + 3(g/L)*sin(a)*w or: 0 = 2 dw/dt + 3(g/L)*sin(a) Since dw/dt = (d/dt)da/dt = a'' 0 = 2 a'' + 3(g/L)*sin(a) ~ 2*a'' + 3(g/L)*a so it boils down to: 0 = a'' + (3g/2L)*a so the angular frequency is w = sqrt(3g/2L) and the period is: T = 2pi/w = 2pi/sqrt(3g/(2L)) = 2*pi*sqrt(2L/(3g)) For L = 0.5 m, T = 2*pi*sqrt(1/(3*9.8)) = 1.159 s b) For a simple pendulum of length 1 m, T = 2*pi*sqrt(1/9.8) = 2.007 s The difference is (2.007-1.159)/1.159 = 0.732 = 73.2%