Statics on an inclined plane

[fedecolo]

Homework Statement

On an incline with slope angle α there lies a cylinder with mass M, its axis being horizontal. A small block with mass m is placed inside it. The coefficient of friction between the block and the cylinder is μ; the incline is nonslippery. What is the maximum slope angle α for the cylinder to stay at rest? The block is much smaller than the radius of the cylinder.

Homework Equations

The Attempt at a Solution

Well, I'm quite sure that I must consider the torques of M and m (maybe the origin of my XY plane can be the contact point between the cylinder and the incline). In this case the torque $T_{M}= R M g sin \alpha$ but how can I write the equation of $T_{m}$ ?
And is it correct to think that the two torque are opposite so until $T_{m} > T_{M}$ the cylinder will stay at rest?

 

[TSny]

1 In this case the torque $T_{M}= R M g sin \alpha$ but how can I write the equation of $T_{m}$ ?

If a line from the center of the cylinder to the small mass m makes an angle $\theta$ to the vertical, can you express the lever arm for the torque $T_m$ in terms of $\theta$ and $\alpha$?

And is it correct to think that the two torque are opposite

Yes

so until $T_{m} > T_{M}$ the cylinder will stay at rest?

The cylinder stays at rest as long as the two torques have equal magnitudes. Also, you don't want the small mass to slip.

 

[CWatters]

Perhaps remember that the slope of the curve on which mass m sits is orthogonal to the line between the centre point and m.

 

[fedecolo]

If a line from the center of the cylinder to the small mass m makes an angle $\theta$ to the vertical, can you express the lever arm for the torque $T_m$ in terms of $\theta$ and $\alpha$?.

Yes, but I don't know how to relate angle $\theta$ with $\alpha$

Because I have a isoscel triangle ( 2 side of length $R$ ). Any hint?

 

[fedecolo]

Is it possible that $\theta = (90 - \alpha )$

 

[CWatters]

θ is measured with respect to YR and YR is orthogonal to the plane X.

 

[TSny]

To find the lever arm for $mg$, it might help to identify the angle shown with the question mark below

I was thinking of $\theta$ as being the angle shown in red below

 

[haruspex]

but I don't know how to relate angle θ with α

They are related through the torque equation.
How does θ relate to μ?
(You will find it easier to work with TSny's definition of θ, as being the angle to the vertical; but either will work.)

 

[fedecolo]

To find the lever arm for $mg$, it might help to identify the angle shown with the question mark below

The angle ? is $\alpha$

How does θ relate to μ?

Considering TSny's $\theta$ angle $\mu = tan \theta$ and m doesn't slide.

 

[haruspex]

The angle ? is $\alpha$
Considering TSny's $\theta$ angle $\mu = tan \theta$ and m doesn't slide.

Right.
Next, can you find the horizontal distance from the point of contact with the slope to the mass ? I see you already have the distance to the cylinder's mass centre line.

 

[fedecolo]

But in order to find it I need the distance of the mass from my origin of axys XY (?)

 

[haruspex]

But in order to find it I need the distance of the mass from my origin of axys XY (?)

what is the horizontal distance from the mass to the centre of the cylinder?

 

[fedecolo]

Is $R sin \theta_{1}$ with $\theta_{1}$ the angle between Y axe and the mass

 

[haruspex]

Is $R sin \theta_{1}$ with $\theta_{1}$ the angle between Y axe and the mass

Right, and you know the horizontal distance from the point of contact to the cylinder's mass centre, so...

 

[fedecolo]

Right, and you know the horizontal distance from the point of contact to the cylinder's mass centre, so...

No because I don't know $\theta_{1}$

 

[TSny]

what is the horizontal distance from the mass to the centre of the cylinder?

Is $R sin \theta_{1}$ with $\theta_{1}$ the angle between Y axe and the mass

Does $\theta_1$ represent your choice of $\theta$ or my choice of $\theta$? (I'm not sure if the Y axis here is the axis perpendicular to the slope or perpendicular to the ground.)

Keep in mind that haruspex asked for the horizontal distance between $m$ and the center of the cylinder.

 

[fedecolo]

My $\theta_{1}$ is different from your choice of $\theta$

Ohh well, the horizontal distance is $R cos \alpha$

But I have to find the torque with respect to my origin of axys, haven't I?

 

[haruspex]

My $\theta_{1}$ is different from your choice of θ

Then I misunderstand how you are defining it. You wrote that it was the angle between the y-axis (i.e. the verical) and "the mass". I took that to mean the radius that leads to the mass. That would make your angle the same as TSny's theta.

Ohh well, the horizontal distance is Rcosα

No, that is the horizontal distance from the centre of the cylinder to the point of contact.

Edit: correction, that is the vertical distance from the centre of the cylinder to the point of contact.

the torque with respect to my origin of axys, haven't I?

Yes. And for that you need to find the horizontal distances from the point of contact to the two mass centres.

 

[RoloJosh16]

$T_m$ gave me $mgR ( \cos \theta - \sin \alpha )$ if $\theta$ is the angle measured from the vertical. So then both torques have to be equal in magnitude an then I obtain a relation between $\alpha$ and $\theta$, but were do I put in $\mu$. $\mu=$ tan $\beta$ is not necessarily true in this exercise if $\beta$ is the slope of the part of the curve where m lies

 

[haruspex]

$T_m$ gave me $mgR ( \cos \theta - \sin \alpha )$ if $\theta$ is the angle measured from the vertical.

Do you mean $mgR ( \sin \theta - \sin \alpha )$?

$\mu=$ tan $\beta$ is not necessarily true in this exercise if $\beta$ is the slope of the part of the curve where m lies

Yes and no. First, you need to identify the way(s) in which the system may become unstable as alpha increases.

 

[RoloJosh16]

Do you mean mgR(sinθ−sinα) mgR ( \sin \theta - \sin \alpha )?

Yes I had made a mistake.

Yes and no. First, you need to identify the way(s) in which the system may become unstable as alpha increases.

Ahh, thanks. The slope of the part where the block lies is $tan \theta$ and you solve it with that.

But is it necessary for the block to be at rest so that the cylinder does not move? I know that then all the equations so far are not true, but anyway.

 

[jbriggs444]

But is it necessary for the block to be at rest so that the cylinder does not move?

It is not clear what situation you are thinking about. Surely if the block is moving, the cylinder will be rocking or rolling. Equally surely, if the block is moving and the cylinder is not rocking or rolling, the block will eventually come to rest. Further, even if the cylinder is rocking or rolling and no other external energy input is provided, the block will eventually come to rest relative to the cylinder.

So please, let us consider the block to be at rest and the system to be in an equilibrium.

 

[RoloJosh16]

It is a little complicated but thanks.

 

[Lnewqban]

... But is it necessary for the block to be at rest so that the cylinder does not move? I know that then all the equations so far are not true, but anyway.

This is a self-balancing system.
We have to determine maximum possible slope.
The cylinder and the little block are not moving initially.

The weights (vertical vectors) of cylinder and block and their horizontal distances (levers) to the fulcrum (cylinder-plane point of contact) balance each other.
Then, the slope of the inclined plane is increased, as slowly as possible.
What happens?