Frictionless half-pipe and rotational motion question

[AJDangles]

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)

 

[AJDangles]

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.

 

[Doc Al]

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!)

 

[AJDangles]

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.

 

[Doc Al]

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:
ƩF_x = ma_x
mg cosθ = ma_x

So... a_x = g cosθ

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

 

[AJDangles]

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:
ƩF_x = ma_x
mg cosθ = ma_x

So... a_x = 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: