Rotation and varying friction coefficient

[Saitama]

Homework Statement

(see attachment)

Homework Equations

The Attempt at a Solution

I am not able to understand the question and build a scenario in my mind. The question asks the distance traveled when the cylinder starts slipping. I can't think of the situation when the cylinder "slips". I am not sure which equations to start with and i suppose this question involves some integration too.

 

[Doc Al]

How much friction force is required to prevent slipping?

 

[Saitama]

How much friction force is required to prevent slipping?

I am not sure but do i have to make the equations for torque and the forces to find the frictional force?

 

[Doc Al]

I am not sure but do i have to make the equations for torque and the forces to find the frictional force?

Yes. Apply Newton's 2nd law for rotation and translation.

 

[Saitama]

Yes. Apply Newton's 2nd law for rotation and translation.

Let the mass of cylinder as M, radius R, friction force f, α as angular acceleration, θ as angle of inclination, and a as the linear acceleration, we have
fR=\frac{MR^2}{2}α
Since, the cylinder does not slip, the equation α=a/R is applicable, so
f=Ma
Now,
Mg\sin(\theta)-f=Ma
Mg\sin(\theta)=2Ma
a=\frac{g\sin(\theta)}{2}
Therefore, the friction force required to prevent slipping is
f=\frac{Mg\sin(\theta)}{2}
Is this correct?

 

[Doc Al]

Let the mass of cylinder as M, radius R, friction force f, α as angular acceleration, θ as angle of inclination, and a as the linear acceleration, we have
fR=\frac{MR^2}{2}α

Good.

Since, the cylinder does not slip, the equation α=a/R is applicable, so
f=Ma

Redo that step and all that follows.

But you're on the right track. Once you have the correct expression for the friction force, ask yourself what minimum value of μ is required to provide that force.

 

[Saitama]

Redo that step and all that follows.

Oops, made a small mistake there.
The f would be
f=\frac{Ma}{2}
Solving using the same method as before, i get
f=\frac{Mg\sin(\theta)}{3}

...ask yourself what minimum value of μ is required to provide that force.

I still can't get the meaning of the problem. Doesn't the cylinder starts slipping the instant it is released or does "slipping" have a different meaning here?

 

[Doc Al]

Oops, made a small mistake there.
The f would be
f=\frac{Ma}{2}
Solving using the same method as before, i get
f=\frac{Mg\sin(\theta)}{3}

Good.

I still can't get the meaning of the problem. Doesn't the cylinder starts slipping the instant it is released or does "slipping" have a different meaning here?

As long as the surfaces are capable of providing the needed friction force (which you have just calculated), there will be no slipping. But on this surface, the value of μ (and thus the maximum available static friction) decreases with distance down the incline. At some point the surfaces will not be able to provide the needed friction and the cylinder will begin slipping. Find that point.

 

[ehild]

Oops, made a small mistake there.
The f would be
f=\frac{Ma}{2}
Solving using the same method as before, i get
f=\frac{Mg\sin(\theta)}{3}



I still can't get the meaning of the problem. Doesn't the cylinder starts slipping the instant it is released or does "slipping" have a different meaning here?

The cylinder starts rolling when released if the static friction is enough. Remember the force of static friction ≤ μ F_N (F_N is the normal force).

ehild

 

[Saitama]

As long as the surfaces are capable of providing the needed friction force (which you have just calculated), there will be no slipping. But on this surface, the value of μ (and thus the maximum available static friction) decreases with distance down the incline. At some point the surfaces will not be able to provide the needed friction and the cylinder will begin slipping. Find that point.

The cylinder starts rolling when released if the static friction is enough. Remember the force of static friction ≤ μ F_N (F_N is the normal force).

ehild

Thanks you both for the explanation, i have got the right answer.

To find that point, i equate the friction force provided by the surface to the friction force required to prevent slipping.
μN=\frac{Mg\sin(\theta)}{3}
Here N is the normal reaction due to the inclined plane and is equal to Mgcos(θ).
\frac{2-3x}{\sqrt{3}}Mg\cos(\theta)=\frac{Mg\sin(\theta)}{3}
Solving, i get
x=\frac{1}{3}

Thanks once again!

 

[Doc Al]

Good!

 

[Tanya Sharma]

Here is my attempt at the problem and i am getting an incorrect result ...Kindly check my work and let me know where am i making mistake

Since at the point of slipping maximum static friction acts

f = μN

where N = Mgcosθ

Thus f =μMgcosθ

but at the same time ,

f = \frac{Ma}{2}

So, \frac{Ma}{2}=μMgcosθ
a=2μgcosθ

now θ =60°

a=μg
μ =\frac{2-3x}{\sqrt3}

a=(\frac{2-3x}{\sqrt3})g

Writing a=v\frac{dv}{dx}

v\frac{dv}{dx}=(\frac{2-3x}{\sqrt3})g

vdv={(\frac{2-3x}{\sqrt3})g}dx

\int_{0}^{v}vdv=\int_{0}^{x}{(\frac{2-3x}{\sqrt3})g}dx

\frac{v^2}{2}=\frac{2}{\sqrt3}gx - \frac{3}{2\sqrt3}g{x^2}

Hence ,v^2=\frac{4}{\sqrt3}gx - \frac{3}{\sqrt3}g{x^2} (1)

Now, we will apply Conservation of energy

When the cylinder has moved by a distance x along the incline ,then

Loss in potential energy =Gain in Kinetic energy

Since rolling without slipping occurs,we can put v=ωR and I = \frac{MR^2}{2}

mgxsinθ = \frac{1}{2}Mv^2 + \frac{1}{2}Iω^2

Solving ,we get v^2=\frac{2}{\sqrt3}gx (2)

Equating (1) and (2) we get x=2/3 which is incorrect.

Where am i getting it wrong?

 

[ehild]

You can not apply conservation of energy.

ehild

 

[Tanya Sharma]

You can not apply conservation of energy.

ehild

why can't we apply conservation of energy?? friction is not doing any work since the cylinder rolls without slipping ...energy is definitely conserved till the point cylinder starts to slip

 

[ehild]

Well, yes, you are right... The energy is conserved.

You used the maximum static friction to calculate the speed. a=2Fs/m, but Fs is not μgsosθ during rolling. It reaches that value only at the end when the cylinder starts to slip.
You can get the acceleration a from the kinematic equations for translation of the CM and rotation about the CM.

ehild

 

[Tanya Sharma]

Well, yes, you are right... The energy is conserved.

You used the maximum static friction to calculate the speed. a=2Fs/m, but Fs is not μgsosθ during rolling. It reaches that value only at the end when the cylinder starts to slip.
You can get the acceleration a from the kinematic equations for translation of the CM and rotation about the CM.

ehild

Okay...Then what should I equate 'a' (accelerataion) with ? Kindly let me know how then will i be able to calculate the speed 'v' at distance 'x' .I have showed you the complete working ...

 

[ehild]

You do not need v really. But if you want it, find the acceleration first and integrate. What equations do you have for the motion of the cylinder?

ehild

 

[Tanya Sharma]

For translation motion
N=Mgcosθ
Mgsinθ - f = Ma

For rotational motion

fR = Iα

For rolling without slipping α = \frac{a}{R}

Putting these values , we get

a=\frac{2}{3}gsinθ

Integrating , we get v^2=\frac{2}{\sqrt3}gx which is the same result which we get from applying conservation of energy

What then is the condition mathematically , which we can apply in finding when the cylinder starts to slip ?

 

[ehild]

Well, a can be determined from conservation of energy. You know a, then you can find the force of static friction. What is it?

ehild

 

[Tanya Sharma]

Well, a can be determined from conservation of energy.

How can we find a from conservation of energy ?

You know a, then you can find the force of static friction. What is it?

Okay...then we are on the same lines as what has been done earlier by Pranav...i.e equating \frac{Mgsinθ}{3} =μN .Isnt it??

 

[ehild]

yes, it is. At the brink of slipping, Fs=μN.

As for acceleration, you can find v from conservation of energy, and then you can differentiate... A bit complicated, and has no sense to do it now. The velocity was not needed at all.

ehild