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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top