So let's say the mass is at rest at the highest position, and is just about to fall. So its potential energy would be say mg(51) instead of mg(50)? And at the zero reference point (ground), its potential energy is still mg(1)?Doc Al said:
If you measure PE from ground level, when the object is resting on the ground its PE = mgd (say). Then when it's at the highest point, resting on top of the track at height h, its PE would be mg(h+d). For some things it doesn't matter, since the change in PE will still be mgh.Sho Kano said:
I see, I just find it weird that its potential energy would be non-zero at ground level. When would one want to use the center of mass to calculate PE?Doc Al said:
Realize that the zero level is arbitrary. Only changes in gravitational PE have meaning.Sho Kano said:
Depends on how accurate you want to be and what you are asked to calculate.Sho Kano said:
In the classic problem, where it asks to calculate the energy needed to get pass the loop, is this required? I'm having trouble thinking up with situations where it would want to be done. What is the size difference to make this matter?Doc Al said:
Think about it. What's the PE initially? What's the PE of the object when it passes the top of the loop? (Note that the object is now below the track.) Since centripetal acceleration is involved, what radius will you use?Sho Kano said:
It depends how accurate you want to be. Many versions of the problem don't even give the dimensions of the object, so you are expected to ignore it.Sho Kano said:
For without accounting for the center of mass:Doc Al said:
OK.Sho Kano said:
How did you arrive at this?Sho Kano said:
Sure. As to how much it matters, that depends on how x compares to r.Sho Kano said:
mg(y+x) = mg(2r-x) + 0.5mrgDoc Al said:
Check that 0.5mrg term. Recall my comment about what radius to use.Sho Kano said:
0.5mrg should be 0.5m(r-x)g?Doc Al said:
Right.Sho Kano said:
That will certainly make things more interesting!CWatters said:
The center of mass is used to determine the stability and balance of an object. In the case of a loop de loop, the center of mass must be located within the loop in order for the object to maintain its circular motion without falling off the track.
The mass and distribution of weight within the object, as well as the velocity and radius of the loop, all affect the location of the center of mass and whether the loop de loop can be successfully completed.
No, the center of mass must be within the loop in order for the object to maintain its circular motion without falling off the track. If the center of mass is outside of the loop, the object will not have enough centripetal force to complete the loop successfully.
The center of mass can be manipulated by adjusting the weight and distribution of weight within the object. By making the center of mass closer to the bottom of the object, it will have a lower center of gravity and be more stable during the loop de loop.
If the center of mass is not taken into account, the object may not have enough centripetal force to complete the loop and could potentially fall off the track. This could result in damage to the object or injury to the person performing the loop de loop.