A Cylinder Rolling in a V-Groove

[ThEmptyTree]

Homework Statement

A cylinder of mass m and radius $R$ is rotating in a V-shaped groove with a constant angular velocity $\omega_0$. The coefficient of friction between the cylinder and the surface is $\mu$.
What external torque must be applied to the cylinder to keep it rotating at a constant angular velocity? Express your answer using some or all of the following variables: $m$ for the mass of the cylinder, $g$ for the gravitational acceleration, $R$ for the radius of the cylinder and $\mu$ for the coefficient of friction.

Relevant Equations

$\vec{\tau}=\vec{r}\times\vec{F}$

I am confused because according to my solution the disk is already rotating at constant angular velocity.
I have written the translational equilibrium on the horizontal and vertical component:
$N_1$ and $f_2$ will have a positive horizontal contribution, while $N_2$ and $f_1$ will have a negative contribution:
$$\frac{1}{\sqrt{2}}N_1-\frac{1}{\sqrt{2}}N_2+\frac{1}{\sqrt{2}}\mu N_2 - \frac{1}{\sqrt{2}}\mu N_1=0$$
$$\Rightarrow N_1=N_2=N$$
All of the forces except gravity will have a positive vertical contribution:
$$\frac{1}{\sqrt{2}}N_1+\frac{1}{\sqrt{2}}N_2+\frac{1}{\sqrt{2}}\mu N_1+\frac{1}{\sqrt{2}}\mu N_2-mg=0$$
$$\Rightarrow N=\frac{1}{(1+\mu)\sqrt{2}}mg$$
The current torque will only come from $f_1$ and $f_2$:
$$\vec{\tau}=R\mu N_2\hat{k}-R\mu N_1\hat{k}$$
... which simplifies because $N_1=N_2$ !

What am I doing wrong?

 

[ThEmptyTree]

Nevermind, $f_2$ is acting in the opposite direction.

 

[haruspex]

You have assumed the groove is a right angle. The question statement you posted does not specify.

 

[ThEmptyTree]

You have assumed the groove is a right angle. The question statement you posted does not specify.

It actually is!

This is their drawing, but they did not specify in the problem.

By the way, I have got $\vec{\tau^\text{ext}}=\frac{2\mu^2}{2+\mu^2}mgR\sqrt{2}\hat{k}$

Hope it's good!

A sincere thanks for being along in my Newtonian physics journey @haruspex! You are an awesome man. Got only 2 more chapters left and I'm done with the physics course (I took a break from it during school).

 

[haruspex]

It actually is!

View attachment 291498

This is their drawing, but they did not specify in the problem.

By the way, I have got $\vec{\tau^{ext}}=\frac{2\mu^2}{2+\mu^2}mgR\sqrt{2}$

Hope it's good!

A sincere thanks for being along in my Newtonian physics journey @haruspex! You are an awesome man. Got only 2 more chapters left and I'm done with the physics course (I took a break from it during school).

You should try to check answers by considering special cases.
If $\mu=1$, one of the normal forces goes to zero, the friction on the other slope being sufficient. It's easy to what the torque would be in that case, and it is not $\frac{2\sqrt 2}3mgR$.
if you can't find your error, please post your working.

 

[ThEmptyTree]

I got these relations from the translational equilibrium:
$$\frac{N_1}{N_2}=\frac{1+\mu}{1-\mu}$$
$$(1+\mu)N_1+(1-\mu)N_2=mg\sqrt{2}$$
Solving for $N_1,N_2$:
$$N_1=\frac{1+\mu}{2+\mu^2}mg\sqrt{2}$$
$$N_2=\frac{1-\mu}{2+\mu^2}mg\sqrt{2}$$
Torque from frictions:
$$\vec{\tau_{f_1}}=-R(\mu N_1)\hat{k}$$
$$\vec{\tau_{f_2}}=-R(\mu N_2)\hat{k}$$
Total torque from frictions:
$$\vec{\tau_f}=-\frac{2\mu}{2+\mu^2}mgR\sqrt{2}\hat{k}$$
Still not good...

 

[ThEmptyTree]

@haruspex Solved it. If $\mu=1$ the additional torque needed is
$$\vec{\tau^\text{ext}}=\frac{1}{\sqrt{2}}mgR$$.
In general
$$\vec{\tau^\text{ext}}=\frac{\mu}{1+\mu^2}mgR\sqrt{2}$$
... which respects the particular situation of $\mu=1$.

 

[haruspex]

@haruspex Solved it. If $\mu=1$ the additional torque needed is
$$\vec{\tau^\text{ext}}=\frac{1}{\sqrt{2}}mgR$$.
In general
$$\vec{\tau^\text{ext}}=\frac{\mu}{1+\mu^2}mgR\sqrt{2}$$
... which respects the particular situation of $\mu=1$.

Well done.

 

[Justforthisquestion1]

Hi I am currently working on the same direction and i find the solution baffling. If someone could please provide an easy to understand solution to why the 2 normal forces do not have the same size? Because from intuition, that would have been my first guess.

 

[berkeman]

Welcome to PF.

Hi I am currently working on the same direction and i find the solution baffling. If someone could please provide an easy to understand solution to why the 2 normal forces do not have the same size? Because from intuition, that would have been my first guess.

I only skimmed the thread, but my guess is that the rotation tends to make the cylinder "climb" up one side, which increases the Normal force for that side and decreases it for the other side.

 

[Lnewqban]

Hi I am currently working on the same direction and i find the solution baffling. If someone could please provide an easy to understand solution to why the 2 normal forces do not have the same size? Because from intuition, that would have been my first guess.

Welcome, @Justforthisquestion1 !

Post #5 may contain the answer to your question.
If you can imagine the assembly as a rack-pinion mechanism, you could visualize the pinion climbing up one side of the V grove and separating itself from the other.

 

[Justforthisquestion1]

Welcome, @Justforthisquestion1 !

Post #5 may contain the answer to your question.
If you can imagine the assembly as a rack-pinion mechanism, you could visualize the pinion climbing up one side of the V grove and separating itself from the other.

View attachment 322642

Right! Normaly the cylinder would climb up the side! And to counterbalance that the normal force on this side needs to be larger! Thank you both!