Physics Help Forum C.G. and stability

 Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum

 Nov 12th 2014, 09:17 AM #1 Junior Member   Join Date: Nov 2014 Posts: 3 C.G. and stability "Lower the cg higher the stability." This is what we are taught in school. But iam little bit confused. Reason for above statement is stated as raising height of cg will increase torque due to its weight that causes body to be unstable. I understand. But isnt that raising height of a body also increases its moment of inertia(resistance to angular acceleration) about point of rotation(contact point between body and ground, in this case) ,because mass of body gets concentrated farther from point about which it rotates? Since,I=Mr^2, I depens on square factor of r, while torque depends simply on r. This implies increase in 'I' is more than increase in torque for certain rise in cg of body. Doesn't this mean stability infact increase with increase in height of cg of body? Might be horribly wrong but please help.
 Nov 12th 2014, 12:26 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 Raising the CG means more torque is applied at the base given the same force acting on the mass. The object reaches instability when that torque exceeds the limit that the base can withstand. For an object like a car if the CG is located at the center the torque that it can withstand before tipping is mgb/2, where b is the distance between the inside and outside wheels. Note that this does not depend in I. Raising the height of the CG causes the torque to be increased, so the object reaches instability at a lower value of force being applied at the mass. For the case of a car in a turn this means that he car will turn over at a lower value of velocity, or following a curve with a greater radius. The fact that I is larger if the CG is raised merely means that once the object starts to tip over it will take longer to hit the ground.
 Nov 13th 2014, 08:45 AM #3 Junior Member   Join Date: Nov 2014 Posts: 3 Thanks ChipB. I wanna know if there is any tern for max. torque base of a body can withstand and also factors on which it depends.
Nov 13th 2014, 09:24 AM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
 Originally Posted by sudarshanpraz Thanks ChipB. I wanna know if there is any tern for max. torque base of a body can withstand and also factors on which it depends.
Max torque before toppling is based on the geometry of the base and the location of the CG. Consider the moments acting about an edge of the base: there is a torque acting in one direction (say, clockwise) and the resisting force of the object's weight acting as a moment in the opposite direction (counter-clockwise). The magnitude of that resisting moment is equal to the weight of the object times the horizontal component of the moment arm from edge of base to CG. If you consider the weight of the object as a force vector W located at the CG, and the moment arm is a vector R connecting the edge to the CG, then the resisting torque is the vector cross product T=RxW. If the applied torque exceeds this value the object will tip.

Last edited by ChipB; Nov 13th 2014 at 11:11 AM.

 Nov 14th 2014, 01:19 PM #5 Junior Member   Join Date: Nov 2014 Posts: 3 Thanks ChipB.

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