Why is friction neglected in the equation for a rolling cylinder on an incline?

AI Thread Summary
The discussion centers on the neglect of friction in the equation for a rolling cylinder on an incline, highlighting that while friction is essential for rolling without slipping, it is often omitted in simplified physics problems. Participants argue that the problem can be solved using both energy conservation and force analysis, with one noting that static friction does not do work since there is no displacement of the point of contact. The conversation also touches on the importance of understanding the conditions under which energy conservation applies, particularly in rolling motion. Some participants suggest that while the problem simplifies reality, it serves an educational purpose in illustrating core principles. The dialogue emphasizes the balance between theoretical models and practical applications in physics.
ritwik06
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Homework Statement


A cylinder is released from rest from the top of an incline of inclination 'n' and length "L". If the cylinder rolls without slipping what will be its speed at the bottom?

The Attempt at a Solution


This is an example question from my textbook! The solution has been given.
It says
0.5 I \omega^{2}+0.5 mv2=mgl sin n
\omega=v/r

But the thing is that I don't understand, why friction has been neglected in the equation. For rolling on a inclined surface, friction must act!. Isn't it??
 
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Lots of times these sort of problems are simplified, and one assumes a frictionless plane. Since in the problem nothing is mentioned about a coefficient of friction, and it would not be possible to solve with the given information if there was friction, I'd say this is one of those problems. It's not entirely realistic, but makes the problem easier.
 
nicksauce said:
Lots of times these sort of problems are simplified, and one assumes a frictionless plane. Since in the problem nothing is mentioned about a coefficient of friction, and it would not be possible to solve with the given information if there was friction, I'd say this is one of those problems. It's not entirely realistic, but makes the problem easier.

I do not agree with you sir. Here is why.
First of all I can calculate the acting friction even though I am not given the coefficient of friction. The condition of rolling without slipping is sufficient.
If I assume the friction to be f.
mg sin n - f =ma
f*r=I*(a/r)
I=0.5m r2

Solving I get the value of f=(mg sin n)/3
an a=2g sin n/3
now applying simple kinematics:
v after traveling L distance along incline=(4g L sin n/3)^0.5
Which is the same as I get from the energy conservation method(neglecting work on by friction.).

There is something more than meets the eye. Both methods seem right but why has been frictional work neglected in energy conservation method?
 
Let me see if I can explain with another question. Suppose a block sits on an inclined plane so that due to friction it does not slide but just sits there. How much work is being done? (i.e. frictional heat being generated?)
 
If you still have trouble with that now picture my block as a tile on the surface of your rolling cylinder.
J.B.
 
jambaugh said:
Let me see if I can explain with another question. Suppose a block sits on an inclined plane so that due to friction it does not slide but just sits there. How much work is being done? (i.e. frictional heat being generated?)

no heat is generated as negative work done by friction is zero since there is no displacement of the body.

but in my case the centre of mass of the body does move so there should have been negative work done by friction.
 
ritwik, looks like you need to look at conservation of angular momentum. I'm a little rusty but i can point you to a brilliant video lecture where this topic is well explained!
http://uk.youtube.com/watch?v=zLy0IQT8ssk
it appears in the first part of the lecture.

hope that helps
mohan
 
I agree that there must be a static friction (not a dynamic one), otherwise I don't see how the cylinder could roll. Without friction it would simply slide without rolling.
I didn't try but I would attempt to solve the problem with conservation of energy. (I believe the mechanical energy is conserved). If I remember well the formula is E=\frac{mv_{CM}^2}{2}+\frac{I_G\omega^2}{2}. I'm a freshman so don't take all what I said as true. :smile:
 
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