What is moment of inertia? How should I calculate MI of any object?
353956
The moment of inertia is of rotational kinematics, its formula depends on the object, its axis of rotation, and the mass is involved to.  So that is how it differs.\nYou can find information on inertia here:\nhttp://en.wikipedia.org/wiki/Moment_of_inertia#Torque\nand here is a table of them\nhttp://en.wikipedia.org/wiki/List_of_moments_of_inertia
46701
v=u+at\nv^2-u^2=2as\ns=ut+1/2at^2\ns=disance traveled\nu=initial vel.\nv=final vel.\na=accelaration\nt=time
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
818010
Angular momentum is conserved,\n\nAngular momentum is (inertia * angular velocity)\n\nInertia for a disk = .5*mass*radius*radius = 0.5 * 5.6 * .2 * .2\n\nThe initial angular momentum is then: 0.5 * 5.6 *.2 * .2 * 1.8 (note that is in rev/s)\n\nThe inertia for the clay is: 0.5 * 3.1 * .04 * .04\n\nThe final angular momentum of disk and clay is then: 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f   where f is the answer you are looking for.\n\nInitial angular momentum = final angular momentum:\n0.5 * 5.6 *.2 * .2 * 1.8 = 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f or\n\nf = (5.6 *.2 * .2 * 1.8)/((3.1 * .04 * .04) + (5.6 * .2 * .2)) in rev/sec
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
840333
You need to consider conservation of angular momentum\n\nI0*w0 = If*wf\n\nI0 is the initial moment of inertia, If is the final moment of inertia, w0 is the initial angular speed, and wf is the final angular speed.\n\nI = Iss + Io\n\nwhere Iss is the (constant) moment of inertia of the student and the stool, and Io is the moment of inertia of the objects. This general relation applies to both the initial and final moments of inertia. Io can be computed from\n\nIo = 2*m*r^2\n\nwhere m is the mass of the object and r is the distance from the axis of rotation. The factor of 2 accounts for the two different objects. The first equation then becomes\n\n(Iss + 2*m*r0^2)*w0 = (Iss + 2*m*rf^2)*wf\n\nwhere r0 is the initial radius (1.0 m) and rf is the final radius (0.40 m). Solving for the final angular speed you have\n\nwf = (Iss + 2*m*r0^2)*w0/(Iss + 2*m*rf^2)\n\nUsing this expression it is easy to compute the final angular speed.
195717
dba_gregg has missed the point that both are same weight\n\nThe moment of Inertia of a hollow sphere  is 1.6 times greater than the solid sphere and so it rolls down slowly.\n\nThe speed of solid sphere will be 1.09 times the speed of hollow sphere.
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
699905
mass initial = Mi\nmass final = Mf\n\nMass growth = Mf - Mi\nPercent growth = Mf/Mi * 100
276222
If I consider a rectangular box of length L and breadth B and mass M. Then Moment of Inertia of the box is :\n\n          I = M * [(L^2)+(B^2)] / 12\n\n      (Notation : x^2 denotes square of x)\n\n  So, for a cube, put L = B = a (say)\n\n    Then,\n           \n        I for cube = M * (a^2) / 6\n\n    (Notation : x^2 denotes square of x)\n\n I hope that this answer satisfies your query.
643815
angular momentum is given by m*v*r\nso know, we have the units are\nkg*m/sec*meter=kg meter^2 per sec\nits dimensional formula will be ML^2T^-1
970512
According to energy conservation law: m2*g*h=m1*g*h+m1*v^2/2+m2*v^2/2+I*w^2/2.\nm2*g*h is initial potential energy of the system;\nm1*g*h is final potential energy of m1;\nm1*v^2/2 is kinetic energy of m1;\nm2*v^2/2 is kinetic energy of m2;\nI*w^2/2 is kinetic energy of pulley with inertia moment of I & final rotary speed w rad/s^2.\nNow m1=0.46kg, m2=0.5kg, h=0.75m, r=0.05m, t=5s, g=9.8m/s^2.\nAs h=a*t^2/2, then a=2*h/t^2 – linear acceleration of m1 and m2. \nAs v=a*t, then v=2*h/t – linear speed of m1 and m2.\nAngular acceleration of pulley w’=a/r; angular speed w=v/r.\nThus 2*(m2-m1)*g*h=(m1+m2)*v^2+I*(v/r)^2, so I=(extract it please and plug values in).\nT1=m1*(g+a), T2=m2*(g-a). Don’t worry, be happy!
963784
tau = I*(alpha) where tau is the torque\nand alpha is the angular accelleration, \nI is the moment of Inertia = 0.5M*R^2 (for a cylinder)\nTau = r*F = 0.16m*40N \n6.4 = M*(0.0256)*90rad/s\nHence M = [6.4]/(0.0256*90)\nM = 2.78kg
172160
The Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to resist bending. The larger the Moment of Inertia the less the beam will bend.\nThe moment of inertia is a geometrical property of a beam and depends on a reference axis. The smallest Moment of Inertia about any axis passes throught the centroid.
793496
The equation involved is \nT = I * alpha  where T = Torque; I = moment of inertia and alpha = angular acceleration.\nFor this system, \nI = m*L1^2 + m*L2^2\n = m*(20^2 + 80^2) * 10^-4 kgm^2\nInitial torque T = m * g * (80-20) *10^-2 J\nthis gives alpha = T / I = 8.65 rad/s^2.\nLinear acceleration = r * alpha\nHence a1 = 8.65 * 0.2 = 1.73 m/s^2\n          a2 = 8.65 * 0.8 = 6.92 m/s^2
284147
Depends on the shape of the object.\n\n- Represent the centroid with an imaginary coordinates.\n- Divide the object into infinitesimal pieces.\n- Create a statement to represent the moment of the pieces about the centroid.\n- Integrate the moments for the range that covers all the pieces of the object.\n\nThe result of integration should be equal to zero\n\n- Solve the equation.\n\nWish you had given me an object, so i could use that as an example.
708999
Find the inertia.  Find the change in inertia.  Use conservation of momentum to  find the new speed. etc,.