WebAs per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′. It is given that, OO′= After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. WebApr 8, 2024 · Best answer Correct option: (A) [ (7ML2) / 48] Explanation: using parallel axis theorem I = ICM + Mx2 (1) where x is distance of axis of rotation from the centre of mass of rod. x = (L/2) – (L/4) = (L/4) also ICM = (ML2 / 12) from (1) I = (ML2 / 12) + M (L / 4)2 = (ML2 / 12) + (ML / 16)2 = [ (7ML2) / 48] ← Prev Question Next Question →
A rod of length L with linear mass density $\\lambda = kx$ is …
WebApr 11, 2024 · I ′ = I + m a 2 Where, m is mass of rod, a is distance of axis from centre and I is moment of inertia along centre axis. So, from it can be said that axis passing through one of the ends is at distance l 2, from centre, so on substituting this in expression we will get, I ′ = 1 12 M L 2 + M ( L 2) 2 ⇒ I ′ = 1 12 M L 2 + M L 2 4 = M L 2 + 3 M L 2 12 WebApr 9, 2024 · Use the formula for the centre of mass of a rigid body to obtain the coordinate of the centre of mass. A rod is a one-dimensional body, hence the y and z coordinates of the centre of mass will be zero. Formula used: The formula for the centre of mass: X C M = ∫ … nim definition banking
The M.I. of a thin rod about a normal axis through its …
WebApr 11, 2024 · Inertia in context to the perpendicular axis at the centre of the square = Iz = 6ma2 = 20kg−m2 Now, using the perpendicular axis theorem, we have, Iz = Ix + Iy = 2Ix (since square has congruent sides) Ix = 2Iz = 12ma2 Edge of the square is at a distance, 2a from the centre. Using the parallel axis theorem, we have Iedge = Ix +m2a 2 WebMar 31, 2016 · since V = Areal of circular face × length = πR2L, we obtain dm = M πR2L × (πR2.dz) or dm = M L dz ...... (1) Step 1. We know that moment of inertia of a circular disk of mass m and of radius R about its central axis is is same as for a cylinder of mass M and radius R and is given by the equation I z = 1 2 mR2. In our case dI z = 1 2dmR2 ...... (2) nimed wires