Particle on a sliding inclined plane

[Zatman]

Homework Statement

A particle of mass m slides on the surface of a plane inclined at an angle θ to the horizontal. The plane itself has a mass M and is allowed to slide on a horizontal surface (such that θ remains constant). There are no frictional forces.

(i) Write equations of motion for the particle and the inclined plane as viewed from an inertial reference frame.

(ii) By resolving the equations in (i) into a suitable coordinate system, find the acceleration of the particle and that for the inclined plane.

2. The attempt at a solution
Let N₁ be the reaction force of the plane on the particle and vice-versa (by NIII), and N₂ be the reaction of the horizontal plane on the inclined plane. If a is the acceleration of the particle and A is the acceleration of the plane:

m\mathbf{a}=\begin{pmatrix}ma_x\\ma_y\end{pmatrix}=\begin{pmatrix}-N_1\sin\theta\\N_1\cos\theta-mg\end{pmatrix}

M\mathbf{A}=\begin{pmatrix}MA_x\\MA_y\end{pmatrix}=\begin{pmatrix}N_1 \sin\theta\\N_2-N_1\cos\theta-Mg\end{pmatrix}=\begin{pmatrix}N_1\sin\theta\\0\end{pmatrix}

(see diagram - obviously forces are not drawn at correct positions on the plane, but this is irrelevant for the question anyway.).

It's part (ii) I'm having trouble with. Ignoring the y-acc of the plane, there are three equations and four unknowns (a_x, a_y, A_x and N₁). How is transforming to a different coordinate system going to help? Surely a fourth equation is needed. Any hints would be appreciated :)

 

[arildno]

First off, you were asked to view this in a non-inertial reference frame. This was NOT made in order to complicate things for you, but actually to help you. What you are basically lacking is such information that the relative normal velocity between the particle and the block remains a CONSTANT zero.

Now, you have put a really good effort into your attempt here, so as a bonus, I will give you a guide on how this can be done.

1. Choice of non-inertial frame:
We choose the frame in which the inclined plane/block is at rest, so that inclined plane is in equilibrium with the fictitious force -MA_x, the particle -mA_x

2. Equations of motion, plane:
N_1sin(theta)-MA_x=0 (x-direction)
N_2-N_1cos(theta)-Mg=0 (y-direction)

3. Equations for motion, particle:
Here, it is most clever to decompose in the tangential and normal direction, since in the tangential direction, the acceleration component relative to the plane is simply equal to the components from the fictitious force and known gravity, while in the NORMAL direction, the acceleration is KNOWN, namely equal to zero (giving an equation relating N_1, A_x and mg to each other).

This yields 4 equations for the unknowns N_2, N_1, A_x, and a_t, the latter being the relative tangential acceleration of the particle.

---
Now, try to set up these equations, and see if you get it right!

 

[Zatman]

Really sorry, I typed "non-inertial" when I meant "inertial" in part (i). I think your hints might apply to (ii) anyway, so let me try...

 

[arildno]

You can perfectly well solve it in an inertial frame, but then you must relate your accelerations properly, so that the relative normal acceleration between particle and plane is zero.
It is simplest to solve it non-inertially.

Good luck!

 

[arildno]

To set up an appropriate set of unit vectors to use, I suggest:
\vec{t}=-\cos\theta\vec{i}-\sin\theta\vec{j},\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}
Thus, for example,
-mg\vec{j}=mg\sin\theta\vec{t}-mg\cos\theta\vec{n}
-mA_{x}\vec{i}=mA_{x}\cos\theta\vec{t}+mA_{x}\sin\theta\vec{n}

 

[Zatman]

Thanks for you help, arildno. I have set up the following equations as viewed from the non-inertial reference frame of the inclined plane, and, for the particle, in a coordinate system parallel and perpendicular to the inclined plane:

Plane:

M\mathbf{A'}=\begin{pmatrix}N_1\sin\theta-MA_x\\N_2-N_1\cos\theta-Mg\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}

Particle:

m\mathbf{a'}\begin{pmatrix}-mg\sin\theta-mA_x\cos\theta\\N_1+mA_x\sin\theta-mg\cos\theta\end{pmatrix}=\begin{pmatrix}ma_t\\0\end{pmatrix}

Which I then solve to give...

A_x=g\frac{\sin\theta\cos\theta}{\sin^2\theta+\frac{M}{m}}

a_t=-g\sin\theta-A_x\cos\theta

Using a_x=a_t\cos\theta+A_x* I end up with

a_x=-g\frac{\frac{M}{m}\sin\theta\cos\theta}{\sin^2\theta+\frac{M}{m}}

The line * I'm not so sure is correct. Please could anyone confirm that I set up the equations correctly and that the line * is also correct?

 

[arildno]

I agree with that.

Remember that if the block has much less mass than the particle (M<<m), it might quite possibly speed off so quickly that the particle lose contact with it.

Thus, I suggest that you look at the acceleration of the particle in the y-particle, and see if it is ever greater than -g. That essentially will give you a condition for when the assumption of contact between particle and plane is no longer valid.

 

[arildno]

Alternatively, see that N_1, the normal force, is represented as a difference.
However, it must NEVER be negative! Agreed?

 

[Zatman]

arildno, thank you very much for all of your help. :)

Yes, I agree that the normal force cannot be negative!

 

[arildno]

Thus, although I haven't gone through the problem on my own, it looks like:
1. Your basic solution seems sound.
2. But, you should check for whether an unphysicality might occur with negative normal force, and thus add to your given solution a requirement for when the solution in 1 is valid (I assume you can represent that requirement in terms of an inequality for the mass ratio, with an appropriate factor of the angle involved).

I'll check it over for myself some time later.

 

[arildno]

To set up an appropriate set of unit vectors to use, I suggest:
\vec{t}=-\cos\theta\vec{i}-\sin\theta\vec{j},\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}
Thus, for example,
-mg\vec{j}=mg\sin\theta\vec{t}-mg\cos\theta\vec{n}
-mA_{x}\vec{i}=mA_{x}\cos\theta\vec{t}+mA_{x}\sin\theta\vec{n}

Okay, as much to satisfy my own curiosity, I'll solve it in my way here:

Equations are:
N_{1}\sin\theta-MA_{x}=0 (1)
N_{2}-N_{1}\cos\theta-Mg=0 (2)
mA_{x}\cos\theta+mg\sin\theta=ma_{t}(3)
N_{1}+mA_{x}\sin\theta-mg\cos\theta=0(4)
From (1) and (4), we get:
(M+m\sin^{2}\theta)A_{x}=mg\sin\theta\cos\theta
That is:
A_{x}=\frac{mg\sin\theta\cos\theta}{M+m\sin^{2}\theta}(5)
Thus, the requirement N_{1}\geq{0} can be written as:
g\cos\theta\geq{A}_{x}sin\theta
Or, using (5)
(\frac{M}{m}+\sin^{2}\theta)\geq\sin\theta
---------------
That is:
\frac{M}{m}\geq\sin\theta{(1-\sin\theta)}(6)

This is the requirement that comes in addition to to the solution of (1)-(4)

 

[Zatman]

Wow, thanks for the additional details! Really appreciate this!

 

[arildno]

Okay, as much to satisfy my own curiosity, I'll solve it in my way here:

Equations are:
N_{1}\sin\theta-MA_{x}=0 (1)
N_{2}-N_{1}\cos\theta-Mg=0 (2)
mA_{x}\cos\theta+mg\sin\theta=ma_{t}(3)
N_{1}+mA_{x}\sin\theta-mg\cos\theta=0(4)
From (1) and (4), we get:
(M+m\sin^{2}\theta)A_{x}=mg\sin\theta\cos\theta
That is:
A_{x}=\frac{mg\sin\theta\cos\theta}{M+m\sin^{2}\theta}(5)
Thus, the requirement N_{1}\geq{0} can be written as:
g\cos\theta\geq{A}_{x}sin\theta
Or, using (5)
(\frac{M}{m}+\sin^{2}\theta)\geq\sin\theta
---------------
That is:
\frac{M}{m}\geq\sin\theta{(1-\sin\theta)}(6)

This is the requirement that comes in addition to to the solution of (1)-(4)

Now, to be absolutely sure there are no other unphysicalities lurking in this problem, we ought to check also the following inequalities (the global inequality must then be the always most stringent requirement, so that ALL inequalities are fulfilled!):
a_{x}\leq{0} (*)
A_{x}\geq{0}(**)
-g\leq{a}_{y}\leq{0}(***)
Now, from (5), we see that (**) is fulfilled.

--
(3) may be written as:
a_{t}=\frac{mg\sin\theta\cos\theta}{M+m\sin^{2}\theta}\cos\theta+g\sin\theta=\frac{M+m}{M+m\sin^{2}\theta}g\sin\theta
Thus, with in my system of coordinates,
a_{x}=A_{x}-a_{t}\cos\theta
we have, for (*):
\frac{mg\sin\theta\cos\theta}{M+m\sin^{2}\theta}\leq\frac{M+m}{M+m\sin^{2}\theta}g\sin\theta\costheta
yielding the not very interesting inequality:
m\leq{M}+m
We have that a_{y}=-a_{t}\sin\theta
Thus, we must have, (***):
-g\leq{-}\frac{M+m}{M+m\sin^{2}\theta}g\sin^{2}\theta\leq{0}
The left-hand side inequality can then be written as (the right hand automatically fulfilled):
\frac{(M+m)\sin^{2}\theta}{M+m\sin^{2}\theta}\leq{1}
which is automatically fulfilled since M\sin^{2}\theta\leq{M}
---
Thus, we have checked that the system does not exhibit any further potential unphysicalities.

 

[arildno]

The correct interpretation of the derived inequality is that at that value, the system of equations is INCONSISTENT, in that (1) and (4) will clash.

 

[Zatman]

Your dedication to the detail in this question is inspiring, arildno.

 

[arildno]

The dedication has more to do with me having messed up privately a couple of times, and getting angry at the damn problem, I had to kill it thoroughly in public!