Frictionless half-pipe and rotational motion question

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A block slides down a frictionless half-pipe, and the problem involves calculating its velocity, tangential acceleration, radial acceleration, and angular acceleration at various angular displacements. The discussion emphasizes the importance of energy conservation, suggesting that potential energy converts to kinetic energy as the block descends. Participants clarify that radial acceleration is not simply equal to gravity but depends on the tangential velocity and radius. To find angular acceleration, the relationship between linear and angular quantities is explored, with a focus on using free-body diagrams to analyze forces acting on the block. The conversation highlights the need for careful diagramming and understanding of vector components in determining the correct forces and accelerations.
  • #31
Doc Al said:
Yes, the triangle you've drawn is incorrect. (But the direction of the force and tangent line are correct, which is what I was focusing on. Sorry about not pointing that out!) Since you are finding components of the weight, you must draw a new right triangle in which the weight is the hypotenuse.

Sanity check: A vector must always be bigger (or at least equal to) its components. So when drawing a right triangle to find components, the components must be the smaller sides, not the hypotenuse.

Ok. I drew my triangle with the mg. I found the components to be:

x = mgcos30
y = mgsin30

We're looking for the x component as the acceleration, so then we can go: mgcos30 = ma. Cancel out the masses, solve for acceleration.

a = gcos30
a = 8.5units/s^2

Right? (Please say yes... ha ha)
 
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  • #32
AJDangles said:
Ok. I drew my triangle with the mg. I found the components to be:

x = mgcos30
y = mgsin30

We're looking for the x component as the acceleration, so then we can go: mgcos30 = ma. Cancel out the masses, solve for acceleration.

a = gcos30
a = 8.5units/s^2

Right? (Please say yes... ha ha)

I'm thinking this would make sense because at angles of 0 and 180, the acceleration is equal to gravity.
 
  • #33
AJDangles said:
Ok. I drew my triangle with the mg. I found the components to be:

x = mgcos30
y = mgsin30

We're looking for the x component as the acceleration, so then we can go: mgcos30 = ma. Cancel out the masses, solve for acceleration.

a = gcos30
a = 8.5units/s^2

Right? (Please say yes... ha ha)
Right! (or yes, if you prefer!)
 
  • #34
Alright! Last question, I promise: Why is it that the x component gives you the acceleration? I realize that I got the right answer but I'm not so sure as to how I got it. I just assumed it was that way because when you put in cos(0)*9.8 you got 9.8. Which makes sense because the circle is completely vertical at that point.
 
  • #35
AJDangles said:
Alright! Last question, I promise: Why is it that the x component gives you the acceleration?
All you're doing is applying Newton's 2nd law in the tangential direction. For the tangential forces (the x-direction), the only force is the x-component of the weight. So:
ƩFx = max
mg cosθ = max

So... ax = g cosθ

Let me know if this gets at your question. If not, ask again.
 
  • #36
Doc Al said:
All you're doing is applying Newton's 2nd law in the tangential direction. For the tangential forces (the x-direction), the only force is the x-component of the weight. So:
ƩFx = max
mg cosθ = max

So... ax = g cosθ

Let me know if this gets at your question. If not, ask again.

Ok... I think I get it now. Thanks a lot. Just one more picture to make sure my triangle is in the right spot:
 

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