gunterZA 1 Posted May 1, 2017 Report Share Posted May 1, 2017 Hi,I need some assistance understanding moments of inertia. I am doing some review for an upcoming exam, however I am slightly stumped by this question. I have already tried googling "moments of inertia" to try and understand the concept better, but I am having issues knowing when to apply which formulas.I have attached my specific question and the "answer" to the question, hopefully someone can help me understand how to get the answer.----Two identical slender rods of length l and mass m are linked together at 90deg, as shown in figure Q4a (image attached), to form a link in a mechanism. Point B is midway between A and C, and points B, D, E, F and G are equally spaced along the lower link.i) Determine the position of the center of gravity of the linkii) Find the moment of inertia of the link about the point Biii) Find the moment of inertia of the link about the point Div) Find the radius of gyration about the point G----ANSWERS:i) Center of gravity is at Dii) I_{B}=(5/12)ml^{2} kg m^{2}iii) I_{D}=(7/24)ml^{2} kg m^{2}iv) k= sqrt(17/24) l m----*My attempts*i)I know inherently that the centre of gravity is at D. How do i prove it mathematically?----ii)I get the moment of inertial about B like so:I_{B} = (ml^{2})/3 + ((ml^{2}/3) - (ml^{2})/4)But i don't understand why I am supposed to subtract. Is it because the centre of gravity is l/2 below B?----iii)For I_{D} I get the answer by doing:I_{D} = ((ml^{2})/3 - (ml^{2})/4) + ((ml^{2})/(3*16)) + (9ml^{2})/(16*3)Why do I subtract (ml^{2})/4) if that rod is above D?----iv)I'm trying to use I_{G} to find the radius of gyration:So for I_{G}I_{G} = (7/24)ml^{2} - ml^{2}Here I am using the parallel axis theorem, but why do I subtract the (md^{2}) of the AC rod from the Moment of inertia about D? Quote Link to post Share on other sites

kabhay 0 Posted May 19, 2017 Report Share Posted May 19, 2017 As for first part choose point B as origin. Now you know the co-ordinate of centre of mass of rod AC and the co-ordinate of centre of mass of rod BG. Now apply the formula of centre of mass. Quote Link to post Share on other sites

OceanBreeze 378 Posted May 19, 2017 Report Share Posted May 19, 2017 To solve for the centroid, you can pick any point on the object to be the reference axis, but it helps to pick a convenient point. In this case, point G is the most convenient as the entire T can be rotated about it in one plane. Point B would require rotations in two planes. The centroid X_{G }= M1X1 + M2X2 / M1 + M2 We don’t know the masses but that is not a problem because it is a uniform rod, the length can stand in for the mass. X_{G} = L1X1 + L2X2 / L1 + L2X_{G} = (4x4) + (4x2) / 4 + 4 = 3So, the centroid is located a distance of 3 units from point G, placing it at point D. Solving for the Moment of Inertia, (I) is a bit more complicated but still very easy. The basic formula is dI = X^{2} dm , and since dm = M/L dx , dI = M/L X^{2} dx I = M/L ∫ X^{2} dx When integrated between the limits L-h and h, where h is some arbitrary distance along the length, I = 1/3 M [ L^{2} -3Lh + 3h^{2} ]When h = 0 as in the vertical part of the T shape, I = 1/3 ML^{2} When h = L/2, as in the case of the horizontal part of the T , I = 1/12 ML^{2}The total moment of Inertia of the T around the point B is just the sum of the two parts, 5/12 ML^{2} The rest is left as an exercise for the readers Quote Link to post Share on other sites

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