- #1

UniPhysics90

- 16

- 0

Hi,

I do a classical mechanics course, and part of it relates to degrees of freedom.

For a system M of point masses with j constraints, there is N=3M-j DoF

For a rigid body, I know there is 6 DoF (3 translational, 3 rotational).

However, I've tried using drawing the constraints on a 4 point mass system to make it a rigid system. I've done this with 5 constraints (3 rigid rods in blue and 2 fixed angles in green). With 5 constraints the formula gives N=(3x4)-5=7 DoF.

Could someone explain this please?

[PLAIN]http://img152.imageshack.us/img152/7426/degreesoffreedom.png [Broken]

It'd be greatly appreciated. Thanks

EDIT: I think I understand this now, because it's a rigid body, you can work out all other positions in the system itself from the 5 given constraints. All that you can't work out, is the rotational and translational DoF of the system as a whole.

I do a classical mechanics course, and part of it relates to degrees of freedom.

For a system M of point masses with j constraints, there is N=3M-j DoF

For a rigid body, I know there is 6 DoF (3 translational, 3 rotational).

However, I've tried using drawing the constraints on a 4 point mass system to make it a rigid system. I've done this with 5 constraints (3 rigid rods in blue and 2 fixed angles in green). With 5 constraints the formula gives N=(3x4)-5=7 DoF.

Could someone explain this please?

[PLAIN]http://img152.imageshack.us/img152/7426/degreesoffreedom.png [Broken]

It'd be greatly appreciated. Thanks

EDIT: I think I understand this now, because it's a rigid body, you can work out all other positions in the system itself from the 5 given constraints. All that you can't work out, is the rotational and translational DoF of the system as a whole.

Last edited by a moderator: