Block Sliding on Incline

I have a question about the typical block sliding down an incline without friction, but for which the incline has a mass and is also free to slide on a horizontal surface.

I get the feeling that the typical frame of reference attached to the incline is not inertial. It is accelerating. Does that effect the time it would take for the small block to traverse the length of the incline as referenced from that frame or am I under thinking this?

I ask because I solved a problem that had this set up, and it asked how far has the incline moved relative to some fixed frame on the ground. I answered the question correctly (according to the designer of the question ) without considering it ( treating the frame moving with the incline as inertial ). However, I have a nagging feeling that the time taken to traverse the length of the incline as seen from the frame moving with the incline will be effected.

@joe-shmo said
I have a question about the typical block sliding down an incline without friction, but for which the incline has a mass and is also free to slide on a horizontal surface.

I get the feeling that the typical frame of reference attached to the incline is not inertial. It is accelerating. Does that effect the time it would take for the small block to traverse the length of ...[text shortened]... traverse the length of the incline as seen from the frame moving with the incline will be effected.

If we fix a non-inertial coordinate system "1" to the incline, and fix inertial coordinate system "2" to the horizontal surface on which the incline sits I get the following system of equations.

ΣF_y1 = 0 = N - mg*cos( ß ) - m ( x2 )"sin( ß )
The last term is the inertial component of y1 in terms of x2

ΣF_x1 = m( x1 )" = mg*sin( ß ) + m ( x2 )"cos( ß )
Again the last term being the inertial component of x1 in terms of x2.

ΣF_x2 = (M+m)( x2 )" = N*sin( ß )

You can solve for ( x2 )" by combining ΣF_x2 and ΣF_y1

( x2 )" = mg*cos( ß )sin( ß )/[ M + m(1-sin( ß ))]

Then you can substitute the result into ΣF_x1 and after the smoke clears

( x1 )" = g*sin( ß ) + (mg*cos²( ß )sin( ß ))/[ M + m(1-sin( ß ))] = Z

These accelerations are constant, and we can solve for the time it takes the small block to get down the incline of length "L". In particular just solve for t² since we will be subbing that into ( x2 )"

t² = 2L/( Z )

and

( x2 ) = 1/2*( x2 )"t² = 1/2*{mg*cos( ß )sin( ß )/[ M + m(1-sin( ß ))]}{2L/(g*sin( ß ) + (mg*cos²( ß )sin( ß ))/[ M + m(1-sin( ß ))])}

So to give you some numbers in the problem the small block had a mass of 1 kg, the incline had mass of 10 kg, the slope was 30 degrees, and the base of the incline had a length of 220 cm.

So ignoring all this non-inertial frame business I got the incline moves 20 cm by the time the mass reaches the horizontal surface. Taking all this into account ( If I did it correctly) I arrive a 19.55 cm. This was supposed to be a Classical Mechanics "Level 1" problem, so I'm thinking the designer just left out the "all frames are inertial" statement in the problem statement. Or it could be that 20 cm was within the margin of error. There currently is no written solution to the problem, so I cant be sure. But if what I'm doing is correct I'll write one up.

Thanks in advance to anyone who takes the time to go through this with me.

@joe-shmo said
If we fix a non-inertial coordinate system "1" to the incline, and fix inertial coordinate system "2" to the horizontal surface on which the incline sits I get the following system of equations.

ΣF_y1 = 0 = N - mg*cos( ß ) - m ( x2 )"sin( ß )
The last term is the inertial component of y1 in terms of x2

ΣF_x1 = m( x1 )" = mg*sin( ß ) + m ( x2 )"cos( ß )
Again the ...[text shortened]... ect I'll write one up.

Thanks in advance to anyone who takes the time to go through this with me.

In my haste I made some algebraic errors.

( x2 )" = mg*sin( 2ß )/(2*(M+m(1+sin²( ß )))

and

( x1 )" = g*sin( ß ) + mg*sin( 2ß )*cos( ß )/(2*( M+m( 1+sin²( ß ) ) ) )

so, from there just carry it out the same way, and I get

δx2 = 18.3 cm

When I tried to verify this in terms of conservation of energy I found out something that took me by surprise. I was thinking that energy must surely be conserved between the analysis with and without the frame being inertial. Looks like I was wrong...

Then, the energy of the system is simply "PE = mgh" in either case. So in the case of the inertial frame ( no inertial forces) looking at the work done on the block as it moves down the incline we are essentially saying:

mg*sin( ß )*l = 1/2*m*v²

l*sin( ß ) = h

mgh = 1/2*m*v²

So I thought, ok...this is the problem with assuming the frame is inertial. The incline must have kinetic energy too, where is it coming from? After all, we only have "mgh" to go around and its all being used by the block.

So I naturally thought, lets add in the kinetic energy of the incline to be shared Looking at a state when the two separate and move away from each other with their own kinetic energy.

mgh = 1/2*m*(v1)² + 1/2*M*(v2)²

in other words

1/2*m*(v1)² = mgh - 1/2*M*(v2)²

So taking the incline energy into account, the velocity of the block must be lower than the velocity where I didn't take it into account.

1/2*m*(v1)² = 1/2*m*v² - 1/2*M*(v2)²

It is here when comparing the inertial analysis vs non-inertial analysis that I find a contradiction. If we calculate the kinetic energy of the block in the non-inertial analysis, I get:

v² = 2*( g*sin( ß ) + Z² )*l, The term Z² is to indicate that the bit added on to g*sin( ß ) is always positive. This simplifies to:

KE = mgh + m*l*Z²

So, I expected less and I got more. What is going on?!? Luckily Google exists, and you can find out about conservation laws (or in general lack thereof) in non-inertial frames. These kinds of propositions were glazed over, or not addressed in my( admittedly limited ) education ( as far as I remember ).

So Anyhow, this explanation is in case anyone decides to chime in with some knowledge on the subject, to better know where I stand.