
#1
Apr2512, 02:38 PM

P: 64

1) When there is no friction , a wheel with a Vcm = X m/s will only be slipping across the surface at a constant velocity.It won't be rotating because there will not be any forces creating torque.
2) The system ( composed of the rotating object ) always tries to find a balance between Rotational and Translational energy so the friction will change direction if Vcm > ωR ( then it will be opposite to the vector of Vcm in order for the rotational energy to be increased) and when Vcm<ωR then the friction will be in the same direction as Vcm , preventing the rotational speed to increase and increasing Vcm instead. 3) So how can a wheel rotate when Vcm = ωR ?? Since no friction is supposed to exist ?! 4) Let's assume that there is a stationary ball of mass M and Radius R on a rough surface(i.e μς=0.1). We give that ball a small push and as a result its centre of mass moves with Vcm = 2m/s. i) How can we find whether the ball's movements until it stops ? Will it ever stop ? is energy conserved ( if the ball is our system)? What about heat dissipation ? ii) Won't the static coefficient become smaller ( kinetic coefficient) ? What about that ? So many questions about this subject which i can't answer one my own so i am asking for your help ! Please do not delete my post again as this isn't a Homework Assignment! Thank you very much :) 



#2
Apr2512, 03:52 PM

Sci Advisor
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P: 26,167

Hi ZxcvbnM2000!
once it is rolling, there will be no further loss of energy through friction (though there will be through rolling resistance, which essentially is the energy consumed by the continual deformation of the ball) 



#3
Apr2512, 04:15 PM

P: 64

Thank you for your time :) 



#4
Apr2512, 04:19 PM

Mentor
P: 10,813

Question About Rotational And Translational Motion
It has a velocity, but no initial rotation. As all forces are finite, the angular velocity cannot make "jumps" and needs some time to reach the value required for rolling.




#5
Apr2512, 04:25 PM

P: 64

Thank you !!!




#6
Apr2512, 06:12 PM

P: 64

" not true … it can have any angular velocity "
One last thing and i will stop bothering you i promise :p Can you explain that a bit ? Also,Okay i understand that the point that is in contact every time with the surface is NOT moving . I understand the conditions etc but WHAT is making the ball to roll Γ = Ια but Γ is a vector and equal to FR , where is that F then ? Please forgive any spelling/grammar mistakes it's not my first language . Thanks :) 



#7
Apr2512, 06:32 PM

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P: 1,842

If the wheel/disk/ball is sliding relative to the surface, the friction is dynamic friction. The wheel/disk/ball will lose energy because of this. The wheel's kinetic energy changes form to heat energy. The frictional force will have such a direction to v tends toward ω/R, but since the wheel/disk/ball is sliding, it's not there yet. If the wheel/disk/ball is rolling along the surface (but not sliding), then v = ωR, and x = θR. Any residual static frictional force keeps it that way. But since the part of the ball/disk/wheel touching the surface is not moving relative to the surface, no work is done. Remember, ΔW = FΔx. But since the relative Δx = 0, the work done by the frictional force is zero too. The wheel/ball/disk loses no energy due to friction/heat in this case. 



#8
Apr2612, 04:48 AM

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P: 26,167

Hi ZxcvbnM2000!
(just got up ) ie, the wheel will keep the same angular speed as it had at the start if eg it was rolling at 30 mph when the road turned into ice, both the linear speed and the angular speed will stay the same … it will look as if it is still rollling!(however, if the ice is sloping, the linear speed will change, but the angular speed won't ) if α is changing, that is caused by the torque (FR) of the friction force (either backward or forward) FR is the magnitude of the vector R x F (strictly, a pseudovector, since it's a cross product of two vectors), along the axle of the wheel (perpendicular to both F and R) but its not true … the friction from the road on an accelerating car is the only (horizontal) external force … so if it's not doing work, where is all the extra kinetic energy coming from? 



#9
Apr2612, 10:48 AM

P: 64

Thank you all very much :)



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