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

In summary, the problem involves a cylinder rolling without slipping down an incline of length L and inclination n. The solution involves using the condition of rolling without slipping and the conservation of energy and angular momentum. Friction is neglected in the equation, which may seem unrealistic but makes the problem easier to solve. The explanation for this is that friction does not do any work in this situation.
  • #1
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 [tex]\omega^{2}[/tex]+0.5 mv2=mgl sin n
[tex]\omega=v/r[/tex]

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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  • #2
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.
 
  • #3
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?
 
  • #4
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?)
 
  • #5
If you still have trouble with that now picture my block as a tile on the surface of your rolling cylinder.
J.B.
 
  • #6
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.
 
  • #7
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
 
  • #8
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 [tex]E=\frac{mv_{CM}^2}{2}+\frac{I_G\omega^2}{2}[/tex]. I'm a freshman so don't take all what I said as true. :smile:
 

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

What is conservation of energy?

Conservation of energy is a fundamental law of physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another.

Why is conservation of energy important?

Conservation of energy is important because it helps us understand and predict how energy is transferred and used in various systems. It also allows us to make informed decisions about resource management and sustainability.

How is energy conserved?

Energy is conserved through various processes, such as conversion between potential and kinetic energy, heat transfer, and chemical reactions. In a closed system, the total amount of energy remains constant.

What are some examples of conservation of energy in everyday life?

Some examples of conservation of energy in everyday life include: turning on a light switch (electrical energy is converted to light energy), riding a bike (kinetic energy is converted to potential energy when going uphill), and using a microwave (electromagnetic energy is converted to heat energy).

How does conservation of energy relate to the environment?

Conservation of energy is closely tied to environmental conservation. By understanding how energy is conserved and transferred in natural systems, we can make more sustainable choices to minimize our impact on the environment. Additionally, conservation of energy is essential in renewable energy sources, such as solar and wind power.

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