Question

A thin uniform rod (mass = 0.16 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of 1.7 s and an angular amplitude of 4.3 degree . (a) What is the length of the rod? (a) What is the maximum kinetic energy of the rod as it swings?

Answer 1

d = L/2

I = (1/3)*m*L^2

T = 2*pi*sqrt(I/m*g*d)

I / mgd = T^2/4pi^2 = (1.7*1.7)/(4*3.14*3.14) = 0.07327

(1/3)*m*L^2/ m*g*(L/2) = 0.07327

(2/3)*L/g = 0.07327

L = 1.077 m

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w = 2*pi/T = (2*3.14)/1.7 = 3.69 rad/s

A = L*theta = 1.077*4.3*(3.14/180) = 0.081 m

KE max = 0.5*I*w^2 = 0.5*(1/3)*m*L^2*w^2

KE = 0.5*(1/3)*0.16*1.077*1.077*3.69*3.69

KE = 0.421 J