How Does Static Friction Affect Energy Equations in Rolling Motion?

AI Thread Summary
Static friction plays a crucial role in rolling motion by enabling the conversion of gravitational potential energy into both translational and rotational kinetic energy. While static friction itself does no work, it facilitates the rolling motion, allowing the sphere to gain rotational kinetic energy without energy loss in an ideal scenario. The total mechanical energy remains conserved, meaning static friction should not be included as a work term in the energy equation. The confusion arises from the need to understand that while friction aids in energy transformation, it does not directly contribute to energy dissipation in this context. Ultimately, static friction is essential for rolling, but it does not appear as a term in the total energy equation.
240rr
Messages
3
Reaction score
0

Homework Statement


This question isn't for a specific problem. Just knowledge to approach a series of problems.
It concerns a problem where a uniform sphere rolls smoothly down a ramp at incline theta. There is a static frictional force on the ramp. I can go on to find acceleration I am just unsure as to the role of the force of friction in the energy equations.

Homework Equations


E=KE+U
KE=1/2mv^2
KE=1/2Iw^2

The Attempt at a Solution


I'm just Unsure as to whether The force of static friction should be included in the original energy equation or not:

mgh = 1/2mv^2 + 1/2Iw^2 + Fx ?
 
Last edited:
Physics news on Phys.org
Did some digging and found an old post

"Even though no work is done by friction, it does cause energy to be transformed into rotational KE. In the ideal case, there would be no loss in mechanical energy. (Of course, in real life there is rolling friction, deformation, etc., which does dissipate mechanical energy.)"

So is this saying that friction creates the rotational kinetic energy? Do I need to include it in the Total Energy equation then as Fx? Still at a bit of a loss. Don't know why I'm having such a hard time picturing what's happening here.
 
i think as the friction is static therefore no work is done by it, the work done against friction should not be included
 
Thanks for the response. That's what i thought but something my professor told me got me thinking i had to include it in the equation for total energy.
 
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 '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