- #1
rmawatson
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Homework Statement
a particle on a slope with angle theta with no friction. v(0) = 0, x(0) = 0, with coordinates i down the slope and j normal to it.
I am confused about why with the constant velocity forumla I get a different answer to my attempted method.. I can't see what's wrong..I need to find the velocity at "l"
x is the top of the slope of a particle on a smooth surface, with no friction,
v(0) = 0, x(0) = 0
along i direction I am starting with:
mgsin(theta) == ma
gsin(theta) == a
using constant acceleration formula
v^2 = v0^2 + 2a0(x-x0)
v^2 = 0 + 2gsin(theta)(x-0)
v = sqrt( 2gl*sin(theta) )
My original attempt below is wrong, but I can't see why. I want to know what it doesn't work the same.
so from
gsin(theta) == a
Integrating wrt t
gsin(theta)t == v + c
Integrating wrt t again
1/2*gsin(theta)t^2 == x + ct + d
with v(0) == 0 and x(0) == 0
0 = c and d = 0
so if I now plugged in 'l' to the equation for position
1/2*gsin(theta)t^2 == l
and solve for t I get,
t = sqrt[ (2l)/(gsin(theta)) ]
so this is the time at which position == l ?
If I then plug this time into the equation for velocity,
gsin(theta)t == v
gsin(theta)*sqrt[ (2l)/(gsin(theta)) ] = v
not the same as with the constant velocity forumla..
why ? what is wrong with this method
Thanks for any help