Primarygun Posted October 15, 2006 Posted October 15, 2006 I don't know how to find the point on an object where the axis of rotation passes through it. ------------------------------------ Imagine, the above is a thick strip. With only mass 10M on the left and 1M on the right. When it is left falling freely in air, what do you recommend about the rotation of it? I don't know if it rotates around a fixed axis of rotation.
Rocket Man Posted October 15, 2006 Posted October 15, 2006 so you have a 10 unit mass on one side and a 1 unit mass on the other.. basically, you have to find the balance point. suppose you put the thing on a fulcrum, where you balance the see-saw the thing will rotate freely in air. the point it rotates around is called the centre of mass with that, it's pretty self explanitory, the centre of mass is a fixed point within the object. you can make an equation for it, distance1 * mass1 = distance2 * mass2 so the distance from the 1 unit mass to the axis is 10 times the distance from the 10 unit mass to the axis. so you divide the length by 11, (10+1) and place the axis 1 unit distance from the 10 unit mass.
Primarygun Posted October 15, 2006 Author Posted October 15, 2006 Oh the point is the centre of mass. Thank you
swansont Posted October 15, 2006 Posted October 15, 2006 It will only start to rotate if there is a net torque. The torque on the strip will be zero in freefall.
Rocket Man Posted October 16, 2006 Posted October 16, 2006 unless there is torque applied by drag, where the center of drag will move to directly follow the centre of mass. key concept in model rocketry.
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