is is possible to have both rotational and linear inertia?
927274
Yes - roll a ball down a hill. It has linear inertia (the balls center of mass moving down the hill) and rotational inertia (the ball is spinning while it moves down the hill).
175247
Not sure what you mean by linear, but linear acceleration can be described:  acceleration = force / mass\n\nArea of circle = pi r^2\n\nVolume of cube = length x width x height\n\nHope it helps
710706
Gravity, friction and centrifigul.  Don't forget various applications of force = leverage, pulleys.
1101278
Torque is the rotational equivalent of linear force.   Its units are newton metres.   Angular velocity is the rotational equivalent of linear velocity.   Its units are radians per second.
753766
First we determine the total moment of inertia of the system (with everyone on the rim).\n\nPeople: I = m r^2 = (65*150) (96^2) = 8.986 * 10^7\nSpace Station: I = 5.08 * 10^8 (it says 5.08 108 in your question, so I'm gonna assume that this is what you meant)\nSum = 5.679 * 10^8 kg*m^2\n\nIf the apparent acceleration = 1g\nThen, using a = v^2 / r , we can get the linear velocity of the edge\n\n9.8 = v^2 / 96\nv = 30.672 m/s\n\nTo get the angular velocity, use w = v / r \n(w is angular vel.)\n\nw = 30.672 / 96 = 0.3195 radians/sec\n\nSince angular momentum is conserved, we can equate the following:\nI * w (before meeting) = I * w (during meeting)\n\nObtain I (during meeting)\n\nPeople: I = m * r^2 = (65*50) * (96^2) = 2.995 * 10^7\nSpace Station: I = remains the same\nSum = 5.380 * 10^8 kg*m^2\n\nConservation of angular momentum:\nBefore = During\n(5.679*10^8) * (0.3195) = (5.380*10^8 ) * w\nw = 0.337 radians/sec\n\nFind the linear velocity:\nv = w * r = 0.337 * 96 = 32.377 m/s\n\nAcceleration to maintain circular motion:\na = v^2 / r\na = 32.377^2 / 96 = 10.92 m/s^2\n\n(that's 1.11g, if you're wondering)\n\nTada!
323062
Rotational mechanics is a huge field and this space wouldn't be enough to explain it. Neverthless you can see this site for the topics it contains: http://www.batesville.k12.in.us/physics/phynet/Mechanics/RotMechanics/Rotate_Mech_Intro.html\n\nIf you want to know a specific think about Rotational Mechanics then ask that. If you don't know anything about it, then I suggest you buy a book and read about it.
101115
It is depended on the moment and the surface moments if inertia of cross section of cantilever.\n\nMoment is force time distance : M=r*f\n\nand I is used to show surface moments if inertia.\n\nthe more the moment more is the deflection.\nthe more the I less is the deflection.
323244
This is conceivable, but only in a very low/no gravity environment.  Thanks for your thoughtful question.
21718
This question sounds like your teacher/ta/whoever wants you to make 3 assumptions (or they already did and you just didnt mention it):\nThey have the same mass (otherwise he would have to specify the masses of each)\nAir resistance is negligible (pretty common)\nThey all roll without slipping (otherwise they all get to the bottom together)\n\nIf you make these assumptions, the order is based on their respective moments of inertia about the center.  They each move downward at a speed based on their rotation, so the order they reach the bottom will be smallest rotational inertia first. Sphere, disk, ring.
960356
Say your mon asks you to go do your homework. You have initial reluctance...you wanna start getting up to go do it but then you don't quite stand up. Inertia is reluctance of matter to be transformed from one state to another. A little child hits a ball and the ball resists motion. It won't go. However, a bigger boy hits the ball, and at a certaion point uses enough force to get the ball to begin to move. That forces needed to change the ball from static state to dynamic, is your inertia. Or, think of what you need to do to get your ice-cream just begin to melt??  \n\nIt's a probably a good idea you read, study and understand the subject matter, than finding supplied answers for homework. It defeats the purpose of homework. Goodluck
823386
Hi Alan the 1st!\nWe neglect the central aperture in CD, then inertia momentum is I=m*r^2/2 for any homogeneous disk, m=0.017kg, r=0.06m. Torq T=I*w’, w’ is angular acceleration, while w=w’*t, w=22 rad/s, t=0.88s, thus T=m*r^2*(w/t)/2, - plug in the proper numbers, be a good boy!
1093836
Centrifugal force and gravity.
970347
e, because of the Newton's 1st law: "An object at rest will remain at rest unless acted upon by an external and unbalanced force . An object in motion will remain in motion unless acted upon by an external and unbalanced force". Assuming inertial frame of reference.\nThe object can still have angular acceleration if there are two forces cancelling each other but are applied to an object at different points. \n\nIn real life there is always something that will slow down the object even in space.
395176
NO, there is no a fixed pattern that could be considered as "inertial motion"
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
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.
510921
No it does not. Takes more energy to twist them. If it was the same you would have a perpetual motion machine. No such thing. You are not creating energy you are only transforming it.