H FIG. 1: The falling chimney described as a rotating uniform stick. The external forces are the weight of the body applied to the center of gravity, and the constraint force at the base A simple integration, using 0= d= d0=2g sin 8, gives the angular velocity cOS suming that the chimney starts moving from rest and is initially in the vertical direction A further integration of Eg. 4, can lead to g(t) in terms of elliptic integrals We recall that the acceleration in polar coordinates can be written as a=r=(r-re er+ (r8+ 2re)ee, so that, for a point A at a fixed distance r from the origin, it becomes a=aer +agee=-ro er treee For a point at two thirds of the height, r=zH, combining Eqs. 3 and 5 we get aB(r 3H)=3H0= gsin 0, proving that this particular point is the center of percussion of the body, as already mentioned in Sect. I The torque equation allowed us to determine the angular acceleration of the motion in Eq. 3. We can use this result and Newton's second law for the motion of the center of mass (CM) of the whole chimney to determine the unknown force F at the base. The vectorial luation Is mraM=W+Fer e 0 W A H r X Y F FIG. 1: The falling chimney described as a rotating uniform stick. The external forces are the weight of the body applied to the center of gravity, and the constraint force at the base. A simple integration, using .. θ = d . θ dt = d . θ dθ . θ = 3 2 g H sin θ, gives the angular velocity . θ 2 = 3 g H (1 − cos θ), (4) assuming that the chimney starts moving from rest and is initially in the vertical direction. A further integration of Eq. 4, can lead to θ(t) in terms of elliptic integrals. We recall that the acceleration in polar coordinates can be written as a= .. r=(.. r −r . θ 2 )ber + (r .. θ + 2 . r . θ)beθ, so that, for a point A at a fixed distance r from the origin, it becomes a = arber + aθbeθ = −r . θ 2 ber + r .. θbeθ. (5) For a point at two thirds of the height, r = 2 3H, combining Eqs. 3 and 5 we get aθ(r = 2 3H) = 2 3H .. θ = g sin θ, proving that this particular point is the center of percussion of the body, as already mentioned in Sect. I. The torque equation allowed us to determine the angular acceleration of the motion in Eq. 3. We can use this result and Newton’s second law for the motion of the center of mass (CM) of the whole chimney to determine the unknown force F at the base. The vectorial equation is m .. rCM = W + F (6) 4