Conceptual question - rotational+translational motion

  • Thread starter Thread starter rbigongi
  • Start date Start date
  • Tags Tags
    Conceptual Motion
AI Thread Summary
Simultaneous rotational and translational motion can be analyzed independently, as forces applied at different points on identical rods result in varying rotational motions while maintaining the same center-of-mass acceleration. The confusion arises from the distribution of energy between translational and rotational kinetic energy, particularly when forces are applied at different locations. When the same force is applied over the same distance, the work done and energy transferred remain constant, but applying the force over the same time can lead to different outcomes. This distinction explains why two identical objects can experience different total kinetic energies despite receiving the same force. Understanding these principles clarifies the relationship between force, motion, and energy in rotational dynamics.
rbigongi
Messages
1
Reaction score
0
So, we are learning about simultaneous rotational and translational motion. I understand that they can be dealt with independently -- for example, if 2 identical rods have a force applied to them at different points, their rotational motion will differ but they will both have the same center-of-mass acceleration. But I am having a really hard time understanding WHY this is. It seems like if a force was applied to the end of the rod, some of the energy given would go towards spinning the rod, and it would move less (translationally) than if a force was applied at the center of one side of the rod. Can anyone clarify why this is not the case?

Relatedly I am having trouble understanding how 2 identical objects given the same force can end up with different total kinetic energies, due to one of them having a rotational component of KE. It just doesn't make sense to me intuitively.
 
Physics news on Phys.org
welcome to pf!

hi rbigongi! welcome to pf! :smile:

force x time = change in momentum

force x distance = work done = change in energy​

if you apply the same force for the same distance, then yes the energy will be the same

but we usually talk about applying the same force for the same time :wink:

(the same impulse)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top