Understanding Friction in Rotational Motion: Solving a Common Misconception

[coldblood]

Hi friends,
Please help me in solving this problem, I'll appreciate the help.

The problem is as:

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1471110_1461726330721139_1078073816_n.jpg

Attempt:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn2/1486585_1461727017387737_725847173_n.jpg

Thank you all in advance.

 

[haruspex]

I don't think there's any shortcut here. Write out the usual force/torque/acceleration equations, assuming rolling contact and taking the point of application of F to be some height x above the centre of mass. It doesn't say whether the cylinder is uniform, so maybe put I as the Moment of Inertia for now.

 

[coldblood]

I don't think there's any shortcut here. Write out the usual force/torque/acceleration equations, assuming rolling contact and taking the point of application of F to be some height x above the centre of mass. It doesn't say whether the cylinder is uniform, so maybe put I as the Moment of Inertia for now.

Taking torque about the centre point and let the cylinder to be uniform,

F. (x) + f(R) = (I).α
=> F(x) + f(R) = 1/2 MR² .α
=>Fx + fR = MR² α/2 - - - - - - -(1)

Force equation,

F - f = ma (a be the acceleration of c.o.m.) and for no slipping, a = Rα

So, F - f = mRα - - - - - - -- - (2)

 

[haruspex]

Taking torque about the centre point and let the cylinder to be uniform,

F. (x) + f(R) = (I).α
=> F(x) + f(R) = 1/2 MR² .α
=>Fx + fR = MR² α/2 - - - - - - -(1)

Force equation,

F - f = ma (a be the acceleration of c.o.m.) and for no slipping, a = Rα

So, F - f = mRα - - - - - - -- - (2)

So eliminate α to find f as a function of x.

 

[coldblood]

So eliminate α to find f as a function of x.

f = F - [2F(R+x)]/ 3MR²

 

[haruspex]

f = F - [2F(R+x)]/ 3MR²

No, that's dimensionally incorrect. Try it again.

 

[coldblood]

No, that's dimensionally incorrect. Try it again.

Yes, Sorry for this.

The correct one is,

f = F - [F(x + R)]/ 2R ; 0≤x≤R

Here, x can not exceed R.

For x = 0, f = F/2 i.e. translational motion is happening.
For x = R, f = 0 i.e. pure rolling

But for any value of x < R, f will have some positive value. i.e. it will act backwards.

 

[haruspex]

f = F - [F(x + R)]/ 2R ; 0≤x≤R

Still not right. You should get a term like 3Rf.

 

[tiny-tim]

hi coldblood!

Taking torque about the centre point …

I don't think there's any shortcut here.

yes there is …

take torque about the bottom of the cylinder (the point of contact and centre of rotation)

[you can do τ = Iα about a moving point provided

i] it is the centre of mass

or

ii] it is the centre of rotation and the centre of rotation is moving in a straight line parallel to the centre of mass ]​

 

[haruspex]

yes there is …

I merely meant that you can't reason it out without getting into equations (as had been attempted in the OP). Yes, some routes to the equations are distinctly faster than others.

 

[coldblood]

Still not right. You should get a term like 3Rf.

are the equations which I wrote in post 3 right?

 

[coldblood]

Still not right. You should get a term like 3Rf.

Yes I got the correct equation,

3Rf = F (R - 2x)

 

[haruspex]

Yes I got the correct equation,

3Rf = F (R - 2x)

Good. Now, what is the range of possible values of x?

 

[coldblood]

what is the range of possible values of x?

3Rf = F (R - 2x)

x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0

 

[haruspex]

3Rf = F (R - 2x)

x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0

Right. Can you select an answer now?

 

[coldblood]

Right. Can you select an answer now?

Well the confusion which is arising is that I got the function

3Rf = F (R - 2x)

x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0

using the condition of pure rolling, I used a = Rα for the bottom point.

 

[haruspex]

Well the confusion which is arising is that I got the function

3Rf = F (R - 2x)

x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0

using the condition of pure rolling, I used a = Rα for the bottom point.

Right, and that does match one of A, B, C, D. Which one?

 

[coldblood]

Right, and that does match one of A, B, C, D. Which one?

Last one when x = R/2, f = 0, Because during pure rolling contact point does not slip. Hence friction is zero.

 

[haruspex]

Last one when x = R/2, f = 0, Because during pure rolling contact point does not slip. Hence friction is zero.

No, x can be anything from 0 to 2R. You need an answer that's valid for any x. If x is not R/2 there will be a frictional force.

You said friction = 0 was the last choice. It's the third choice of four.

 

[coldblood]

No, x can be anything from 0 to 2R. You need an answer that's valid for any x. If x is not R/2 there will be a frictional force.

You said friction = 0 was the last choice. It's the third choice of four.

I got it. That means Any of the choice could be there. Hence Can't be interpreted?

 

[coldblood]

I got it. That means Any of the choice could be there. Hence Can't be interpreted?

The post which I wrote, Because during pure rolling contact point does not slip. Hence friction is zero.. Is this statement correct?

 

[haruspex]

The post which I wrote, Because during pure rolling contact point does not slip. Hence friction is zero.. Is this statement correct?

No. During rolling the frictional force has a magnitude anywhere from 0 to μ_sN.

 

[ehild]

The friction is zero if the applied force is zero and the cylinder rolls with constant velocity (and angular velocity)

By the way,the problem did not say that it was pure rolling, did it?

ehild

 

[haruspex]

No, x can be anything from 0 to 2R.

Scratch that. I forgot that we defined x to be the height of the force above the centre of the disc. So x is from -R to +R. That makes the frictional force anything from -F/3 to +F.

 

[haruspex]

The friction is zero if the applied force is zero and the cylinder rolls with constant velocity (and angular velocity)

Yes, but not only if. In the context of the OP, it will be zero if F is applied at height 3R/2 from the ground.

 

[ehild]

You are right, it can be zero also in special cases: in this problem, when F is applied at the height 3R/2. But in case of a free rolling object, the static friction is zero. (Of course, it will not roll forever, as there is also the rolling resistance.)

The statement of the OP

Because during pure rolling contact point does not slip. Hence friction is zero.

is wrong.

Correctly: "During pure rolling, the contact point does not slip. Hence friction is static."



ehild

 

[coldblood]

You are right, it can be zero also in special cases: in this problem, when F is applied at the height 3R/2. But in case of a free rolling object, the static friction is zero. (Of course, it will not roll forever, as there is also the rolling resistance.)

The statement of the OP

is wrong.

Correctly: "During pure rolling, the contact point does not slip. Hence friction is static."
ehild

Yes, I got it.

Thank you all for the help.