Center of mass acceleration for an inclined plane and mass m

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SUMMARY

The discussion centers on the acceleration of the center of mass (CoM) for a two-body system consisting of a block on an inclined plane. Participants clarify that while the CoM accelerates vertically due to the motion of the block, the horizontal forces cancel out, resulting in zero horizontal acceleration. Key methods for analyzing this problem include Lagrangian mechanics, Conservation of Momentum, and Newton's Laws in non-inertial frames. The conversation emphasizes the importance of understanding relative motion and the relationship between the two masses in the system.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with Conservation of Momentum and Energy
  • Knowledge of Newton's Laws in non-inertial frames
  • Basic calculus for analyzing motion and forces
NEXT STEPS
  • Learn how to apply Lagrangian mechanics to two-body systems
  • Study the relationship between vertical and horizontal displacements in inclined planes
  • Explore the kinematic equations for relative motion in two-body systems
  • Investigate the calculation of normal forces in non-inertial frames
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the dynamics of two-body systems on inclined planes.

Sam Jelly
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Homework Statement
Mass m is on top of an inclined plane of mass M, let mass m slide down. (All surfaces are frictionless) I’m trying to find the center of mass’s acceleration.
Relevant Equations
(m+M)Acm=Fexternal
FC4C23D1-57DD-4F86-8580-60CD7439F5FF.jpeg

Why is the center of mass acceleration zero? Did I do anything wrong?
 
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Sam Jelly said:
Why is the center of mass acceleration zero?
There is no net external force on the system in the horizontal direction. Although I don't think you properly analyzed it, because in the vertical direction the center of mass is accelerating.
 
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erobz said:
There is no net external force on the system in the horizontal direction. Although I don't think you properly analyzed it, because in the vertical direction the center of mass is accelerating.
All the external forces are in the vertical direction but it all cancels out.
 
Sam Jelly said:
All the external forces are in the vertical direction but it all cancels out.
The vertical coordinate of the combined center of mass ( block and ramp) is accelerating as the small block accelerates down the ramp. What you did is not correct.
 
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erobz said:
The vertical coordinate of the combined center of mass ( block and ramp) is accelerating as the small block accelerates down the ramp. What you did is not correct.
How can I calculate Acm in the y axis? I know that the center of mass is accelerating down in the vertical axis but I don’t know how to calculate the external forces acting on it.
 
Sam Jelly said:
How can I calculate Acm in the y axis? I know that the center of mass is accelerating down in the vertical axis but I don’t know how to calculate the external forces acting on it.
This is a tricky problem because both blocks accelerate. Where did you get this problem?

Have you studied Lagrangian mechanics?
 
I believe applying Conservation of Momentum, Energy are the most straightforward, other methods include examining Newtons Laws in a non-inertial frame for each mass, or the mathematical machinery of the Lagrangian (if you've studied it as @PeroK mentions).

I'll yield any further assistance to those that know each of these methods by heart (I'd have to go through it myself first - if @PeroK says it's tricky - I'm probably underestimating it), as I'm currently cooking Easter dinner!
 
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PeroK said:
This is a tricky problem because both blocks accelerate. Where did you get this problem?

Have you studied Lagrangian mechanics?
It’s a problem from my high school class. We are studying about 2 body systems like reduced mass and center of mass etc. I am confused why all external forces in the vertical axis all cancel out.
 
They don't cancel out the way you think they do or imply they do in your calculations. Imagine a block on a scale. The scale reads ##mg##, now take that block and scale lift it up and drop it, what does the scale read while it's in freefall?
 
  • #10
erobz said:
They don't cancel out the way you think they do or imply they do in your calculations. Imagine a block on a scale. The scale reads ##mg##, now take that block and scale lift it up and drop it, what does the scale read while it's in freefall?
It’s zero. So does that mean my Normal force here is incorrect?
 
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  • #11
Sam Jelly said:
It’s zero. So does that mean my Normal force here is incorrect?
Yes, the normal force on the base of the wedge is incorrect. That little block has some aceleration in the vertical direction between ##0 ~[m/s^2]## and ##g##(freefall) in magnitude.
 
  • #12
I see, is there any way to calculate the normal force?
 
  • #13
Sam Jelly said:
I see, is there any way to calculate the normal force?
There are probably a handful, but none are simple, (how far along you are in you your physics)? Calculus, Momentum, Energy, Newtons Laws in non-inertial frames? I think finding the normal force is one of the more involved you'll find tied to this problem. The experts around here would be better at suggesting steppingstone problem that is at your level of experience than me trying to make one up for you. I think one might be to instead ask you to find what is the velocity of the wedge as a function of the velocity of the block with respect to the slope. Can you think of what principle you might apply to solve that one instead?
 
  • #14
erobz said:
There are probably a handful, but none are simple, (how far along you are in you your physics)? Calculus, Momentum, Energy, Newtons Laws in non-inertial frames? I think finding the normal force is one of the more involved you'll find tied to this problem. The experts around here would be better at suggesting steppingstone problem that is at your level of experience than me trying to make one up for you. I think one might be to instead ask you to find what is the velocity of the wedge as a function of the velocity of the block with respect to the slope. Can you think of what principle you might apply to solve that one instead?
I am familiar with calculus. The actual problem is to find the d(v relative)/dt and the center of mass acceleration. Right now I am trying to solve this problem using the 2 body system.
 
  • #15
Sam Jelly said:
I am familiar with calculus. The actual problem is to find the d(v relative)/dt and the center of mass acceleration. Right now I am trying to solve this problem using the 2 body system.
While you wait for someone to help you with that (I've had a long day - I'm not in the frame of mind to start that tonight). Can you solve the problem I asked (as a test of will ) what is the velocity of the wedge ##u## as a function of the velocity of the block relative to the slope ##\dot s ##?
 
  • #16
Also, if someone is going to help you through this problem it is critical that you learn to use the math formatting language LaTeX Guide It's what we use here to present out mathematics in a clean and organized fashion.
 
  • #17
Sam Jelly said:
I see, is there any way to calculate the normal force?
Not directly. The problem is that downward acceleration of the mass centre is limited by the horizontal inertia. You cannot solve for the two directions independently.
You could look at the kinematic relationship between the vertical and horizontal positions of the small mass (since the horizontal relationship of the two objects is known) and/or consider work conservation.
A sure way, but maybe longer, is to consider the normal force between the objects and apply F=ma to each separately.
 
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  • #18
Sam Jelly said:
Homework Statement: Mass m is on top of an inclined plane of mass M, let mass m slide down. (All surfaces are frictionless) I’m trying to find the center of mass’s acceleration.
Relevant Equations: (m+M)Acm=Fexternal

View attachment 342600
Why is the center of mass acceleration zero? Did I do anything wrong?
The first thing to note is that the small mass has some velocity ##\vec v## relative to the ground; and, it has some velocity ##\vec v'## relative to the block. And, the block has some velocity ##\vec V## relative to the ground.

Imagine what happens in a small time interval ##\Delta t##:

1) The block will have moved ##\vec X = \vec V\Delta t## to the left.
2) The small mass will have moved ##\vec s' = \vec v'\Delta t## down the block.
3) The displacement of the small mass relative to the ground is the vector sum of these two displacements:$$\vec s = \vec X + \vec s'$$4) And, the displacement of the small mass can be expressed in terms of its velocity relative to the ground:$$\vec s = \vec v \Delta t$$Note that by differentating these equations we can see that the acceleration of the mass is the vector sum of the acceleration of the block and the acceleration of the mass relative to the block:$$\vec v = \vec V + \vec v'$$$$\vec a = \vec A + \vec a'$$Any solution will have to deal with the relative motion of the objects. So, I think these equations are your starting point.
 
  • #19
I had a look at this and found a relatively elementary way to do things. Note that in the ground frame:

1) The horizontal displacement of the mass and the block are related.

2) The vertical displacement of the mass is related to the horiziontal displacement. To calculate this, draw a diagram of the starting position and another diagram when the mass has moved some distance down the slope.

3) Express the KE of the system in terms of the vertical displacement of the mass.

4) The PE lost by the system can, of course, be described by the vertical displacement of the mass.

5) Combine and solve those equations to get the vertical acceleration of the mass.

6) Calculate the acceleration of the CoM from that.
 
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