- #1
OskarBillington
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So, this seemed really fun to me until I got stuck.
THE TASK is about an object with mass m, moving in a basic (2D) coordinate system. It is attached to origo (0, 0) by a "rope" with constant length r=5. In position P0(-5, 0) it has the velocity v0=[0, -10]. Hence, the object moves around origo in a circular loop, and the force on m from the rope shall be referred to as C. There is also another (constant) force: F=[10m, 0] (m being the mass: all calculations should be done without units). F could represent a gravitational pull, to ease the comparison to a standard loop. By the terms, the mechanical energy is conserved.
First, you are asked to find the acceleration (expressed as a vector) in points (-5, 0), (0, -5), (5, 0), (0, 5). This is easily done by Newton's 2nd law: ΣF=ma⇒ either m[±v2/r, 0] (in points (-5, 0) and (5, 0), because ΣF=centripetal force=mv2/2) or m[10, ±v2/r] (in the other two points, because ΣFx=F and ΣFy=centripetal force).
FINALLY: You are now asked to find a function a(α)=[ax, ay].
- a is the acceleration vector of m.
- α is the angle, determining the position of m. Preferably, you want this to be the angle from the positive x-axis to the "rope" (counter clockwise).
- The vector function shall describe the acceleration vector of m at any position in the loop.
Good luck
Notes:
It seems that the reason why the second task is hard is because F interferes with Cx. In comparison: in the first task, F has either everything or nothing to do with C. C does of course consist of the pull from the rope, and in most positions also some push/pull from F. I can hardly explain the issue any better (at least in English), but you may anyhow find that the task is quite difficult.
An idea of mine is to define the x-/ y-axes along the velocity vector and the C vector, so that when α changes, F only rotates around m. In this redefined coordinate system: ΣFx will then depend only on the angle, and will simply* be equal to Fx (*still difficult to cover the entire 360° movement with one function). ΣFy will always equal C, again equal to mv2/r - and a function of the size v may be possible to work out. With this x-/y-perspective I suppose you will result in a vector function for a, which would then have to be modified to "rotate" back to work with the x-/y-axis definition in the original task.
I am inexperienced with the English physics terms, hence the poor or incorrect use of terminology. Please inform me on potential for improvement!
THE TASK is about an object with mass m, moving in a basic (2D) coordinate system. It is attached to origo (0, 0) by a "rope" with constant length r=5. In position P0(-5, 0) it has the velocity v0=[0, -10]. Hence, the object moves around origo in a circular loop, and the force on m from the rope shall be referred to as C. There is also another (constant) force: F=[10m, 0] (m being the mass: all calculations should be done without units). F could represent a gravitational pull, to ease the comparison to a standard loop. By the terms, the mechanical energy is conserved.
First, you are asked to find the acceleration (expressed as a vector) in points (-5, 0), (0, -5), (5, 0), (0, 5). This is easily done by Newton's 2nd law: ΣF=ma⇒ either m[±v2/r, 0] (in points (-5, 0) and (5, 0), because ΣF=centripetal force=mv2/2) or m[10, ±v2/r] (in the other two points, because ΣFx=F and ΣFy=centripetal force).
FINALLY: You are now asked to find a function a(α)=[ax, ay].
- a is the acceleration vector of m.
- α is the angle, determining the position of m. Preferably, you want this to be the angle from the positive x-axis to the "rope" (counter clockwise).
- The vector function shall describe the acceleration vector of m at any position in the loop.
Good luck
Notes:
It seems that the reason why the second task is hard is because F interferes with Cx. In comparison: in the first task, F has either everything or nothing to do with C. C does of course consist of the pull from the rope, and in most positions also some push/pull from F. I can hardly explain the issue any better (at least in English), but you may anyhow find that the task is quite difficult.
An idea of mine is to define the x-/ y-axes along the velocity vector and the C vector, so that when α changes, F only rotates around m. In this redefined coordinate system: ΣFx will then depend only on the angle, and will simply* be equal to Fx (*still difficult to cover the entire 360° movement with one function). ΣFy will always equal C, again equal to mv2/r - and a function of the size v may be possible to work out. With this x-/y-perspective I suppose you will result in a vector function for a, which would then have to be modified to "rotate" back to work with the x-/y-axis definition in the original task.
I am inexperienced with the English physics terms, hence the poor or incorrect use of terminology. Please inform me on potential for improvement!
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