- #1
toesockshoe
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Homework Statement
I also need help on a problem. but FIRST, can someone explain to me how radial acceleration relates to gravity and how gravity can be split up into tangential and radial accelerations and how that can be compared to radial acceleration ? I know the basic stuff..like if a object is moving in a circle with constant speed it has radial acceleration, but what happens when gravity comes into play? Look below for a more specific problem oriented question. I was wondering why g is not part of the centripetal acceleration formula, v^2/r.
A small block of mass m slides along the frictionless loop-the-loop track (of radius r) shown in the diagram. What would the minimum height be such that the block goes all the way around the loop?
Here is a visual representation: http://nebula.deanza.edu/~Newton/4A/4AHWSet6.html (its number 2)
Homework Equations
[tex] W= \Delta E [/tex]
KE = 1/2mvf^2=1/2mvi^2
GPE = mgh
The Attempt at a Solution
I chose the system to be the mass and earth. I set the equilibrium point to be the top of the loop.
[tex] W = \Delta E [/tex]
[tex] W_{normal} = \Delta GPE + \Delta KE [/tex]
Work done by nomal force is 0.
[tex] 0 = mg(h-2r)+\frac{1}{2}mv_f^2 [/tex]
solve for h:
[tex] h=\frac{v_f^2}{2g}+2r [/tex]
Now Ill do a F=ma problem to find the velocity.
Let my system be the mass at the top of the loop wit the y direction going down.
[tex] F_g-F_n = ma_c [/tex]
[tex] F_g -ma_c = F_n [/tex]
[tex] mg- \frac{mv^2}{r} = F_n [/tex]
HERE IS MY QUESTIONS: Why is the acceleration the same as uniform circular motion acceleration? Shouldnt gravity also be factored into the acceleration. I was wondering why the acceleration can't be:
[tex] a_c = \frac{mv^2g}{r} [tex]
anyway the rest of the problem is straightforward after and I know the answer... but my main question is why is gravity not in the problem.