# Exercise 1: Inertia moment of a simple pendulum

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1 Exercise : Inertia moment of a simple pendulum A simple pendulum is represented in Figure. Three reference frames are introduced: R is the fixed/inertial RF, with origin in the rotation center and i along the vertical, pointing down; R c, a translation of R in the CoM C of the mass m; the translation vector is t c = [ lc θ ls θ ] T R, with the same origin of R, with i aligned along the the massless bar of length l that connects the rotation center with the mass m. The mass position vector represented in R is r m = l c θ s θ ; r m = l (s θ +c θ ) = l We know from physics that the inertia moment of the point mass with respect to the rotation axis k (coming out from the plane), is ml. WewanttocomputetheinertiamatrixofthemasswithrespecttoO, usingthenotationintroduced in the course, namely Γ. Its general formula is [ r Γ = m I r m (r m )T] dm = Γ xx Γ xy Γ xz Γ yx Γ yy Γ yz B Γ zx Γ zy Γ zz that in our case (a single point mass) becomes Γ = ml c θ s θ c θ s θ c θ s θ = ml c θ s θ c θ s θ c θ s θ Considering only the element Γ zz, we have exactly ml as expected. Now we compute the inertia matrix using the parallel axes theorem. First we notice that the angular momentum does not change if it is represented in R or in R c, h = h c, since the two RF are simply translated. The inertia matrix computed in the CoM of m is the zero matrix, since the mass is a point mass: Γ c = From the parallel axes theorem, we have: [ t Γ = Γ c +m c I t c (t c )T] developing the relation, one obtains Γ = +ml ml c θ s θ c θ s θ c θ s θ

2 that coincides with the inertia matrix computed above. Now we want to compute the inertia matrix with respect to the origin O of R, but represented in the local frame R ; now the mass point vector is represented in R Therefore r ] m[ m I r m (r m )T = r m = l = ml m l = ml ml Exercise : Pendulum with two fixed point masses In Figure, two point masses m m, are connected by a negligible mass bar of length d; the midpoint C of the two connected masses is fixed at the extremity of a pendulum of length l. The two masses are always oriented along the pendulum bar. Therefore the two masses are at a distance l = l d,l = l+d from the pendulum pivot O. We introduce two Rfs, R with origin in O and R with origin in C We want to compute the inertia matrix with respect to the point C, and also with respect to the point to O. The positions of the two masses in R are given by the following geometrical vectors r = l c θ s θ r = l c θ s θ while in R are given by r = r = where r i represents the mass m i in R and r i presents the mass m i in R. The general relations for inertia moments and products are Γ xx = i Γ yy = i Γ zz = i m i (r i +r i3 ) m i (r i +r i3) m i (r i +r i) Γ xy = i m i r i r i Γ xz = i m i r i r i3 Γ yz = i m i r i r i3 where the index i defines the i-th component of the vector.

3 We compute Γ with respect to R origin O, as follows Γ,xx = m l s θ +m l s θ Γ,yy = m l c θ +m l c θ Γ,zz = m l +m l Γ,xy = m l s θc θ m l s θc θ Γ,xz = Γ,yz = Since l = (l d) and l = (l+d), one obtains Γ,xx = m t (l +d )s θ +(m m )ld s θ Γ,yy = m t (l +d )c θ +(m m )ld c θ Γ,zz = m t (l +d )+(m m )ld Γ,xy = m t (l +d ) (m m )ld s θ c θ Γ,xz = Γ,yz = where m t = m +m. If m = m = m, m t = m and having defined (l +d ) = L, we have Γ = m tl s θ m t L s θ c θ m t L s θ c θ m t L c θ m t L Now we compute Γ with respect to R origin C i.e., Γ,xx = Γ,yy = m d +m d = m t d Γ,zz = m d +m d = m t d Γ,xy = Γ,xz = Γ,yz = Γ = m t d m t d Asyoucannotice, the twoinertiamatricesaredifferent; Γ istime-independent, whileγ depends on the angle θ. We compute now some relations between the two inertia matrices. For this purpose, we introduce a fictional reference frame R obtained by a translation of R in O; since the new axes are parallel to the axes of R we can apply the parallel axes theorem and compute the inertia matrix Γ with respect to point O. The inertia moments are ( ) Γ,xx = Γ,xx +m t t y +t z ( ) Γ,yy = Γ,yy +m t t x +t z () ( ) Γ,zz = Γ,zz +m t t x +t y. 3

4 and the inertia products are Γ,xy = Γ,xy m t t x t y Γ,xz = Γ,xz m t t x t z Γ,yz = Γ,yz m t t y t z () since t = [ l ] T, it follows that and i.e., Γ,xx = +m t = Γ,yy = m t d +m t l = m t L Γ,zz = m t d +m t l = m t L. Γ,xy = m t = Γ,xz = m t = Γ,yz = m t = Γ = m t L m t L (3) (4) As a last step we transform Γ into Γ using the relation where Γ = R Γ (R )T R = c θ s θ s θ c θ (R )T = c θ s θ s θ c θ Hence c θ s θ s θ c θ m t L c θ s θ s θ c θ == m tl s θ m t L s θ c θ m t L s θ c θ m t L c m t L θ m t L theta is equal to Γ ; in conclusion we have verified the consistency of the various transformations between inertia matrices. Exercise 3: Pendulum with two rotating point masses In Figure 3, two point masses m,m, are connected by a negligible mass bar, with length d; the bar and the two masses rotate around a pivot in C. This pivot is located at the extremity of a pendulum with length l. The two masses are at the same distance d from the rotation center C. For simplicity we assume m = m = m so that m +m = m t = m. Similarly to the previous exercise, we want to to compute the various inertia matrices. The generalized coordinates q (t) and q (t)are assumed as in Figure 3. In R the two masses are represented by the following geometrical vectors [r ] R = d c s [r ] R = d c s with r i = d 4

5 The rotation matrices between the RFs are R = c s s c R = c s s c and the translations of the origins are [t ] R = l c s [t ] R = d c s therefore, the inertia matrix with respect to the pivot point C expressed in the RF R is: Γ c = ( m i r i I r i (r i ) T) = m t d m t d c c s c s s i = m t d s c s c s c recalling that c i = s i and s i = c i. Nowwecompute Γ c andafterthat, successivamenteγ c = R Γ c R, in orderto verifythe equality of the two results. We observe that the change in the reference frame has nothing to do with the point with respect to which the angular moment is computed, that, in this present case, remains the point C. Now we express the position of the two masses in R, as with r i = d. [r ] R = R [r ] R = c T s s c c s d = d [r ] R = R [r ] R = c T s s c c s d = d Hence d Γ c = m d I [ d ] d T +m d I [ d ] T = m t d and Γ c = R Γ c R = c s s c The two results are equal as expected. Γ c c s s c = m t d c c s c s s 5

6 Figures Figure : Esercizio : a simple pendulum. Figure : Esercizio : a pendulum with two point masses. 6

7 Figure 3: Esercizio 3: Pendulum with two rotating point masses. 7

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