
#1
Nov2805, 12:54 AM

P: 214

The textbook solution for a pipe rolling down an incline is [tex]a = \frac{1}{2} g \sin \theta [/tex]. Putting in theta = 0 gives a = 0. Does this imply that a ball rolling along a level surface will never stop?




#2
Nov2805, 01:01 AM

PF Gold
P: 7,125

It means that the object will not accelerate on a level plane if gravity is the only force.




#3
Nov2805, 01:07 AM

P: 214

Gravity is not the only force on an incline. Friction acts as well. And friction acts on a level plane too. So a rolling ball should come to a stop due to friction....




#4
Nov2805, 01:18 AM

HW Helper
PF Gold
P: 1,123

Will a rolling ball ever stop?
the textbook solution was for an "idealized model" with zero friction.
The MODEL has zero friction. Your fault if you try to apply this model to situations where friction is NOT ignorable. You're right, the model has a (hard) surface's Force as well as gravity's Force. And it suggests that the pipe rolling on a level floor will not slow down. Don't try this a home ; you'll soon overreach this model when the pipe hits the kitchen table. 



#5
Nov2905, 11:05 AM

P: 41

You said that this is the formula for a pipe rolling. If it was sliding, then friction would be a problem, but as it is rolling, friction shouldn't be the problem should it?
If I'm right, then the error would be due to some of the kinetic energy of motion being converted to heat, due to the particles of the pipe hitting against the particles in the surface when it is rolling, and drag. So the textbook model was from an idealised model with friction, but with no energy loss as heat or drag. 



#6
Nov2905, 11:14 AM

HW Helper
PF Gold
P: 1,198

There is also the fact that, it is not exactly one point that touches the ground during rolling motion as we assume in the ideal case. There will be an area which is in contact with the ground. Due to this the Normal force does not pass exactly through the center and hence there's a restoring torque which slows down the pipe.



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