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2k
' g # kv ( (1)
(2) where the origin is taken to be the point at which v $ v0 so that the initial condition is
x ) v $ v0 * $ 0 . Thus, the distance from the point v $ v0 to the point v $ v1 is s ) v0 + v1 * $ 212. (3) The equation of motion for the upward motion is 2
! g # kv0 "
1
log %
2&
2k
' g # kv1 ( m d2 x
$ # mkv 2 # mg
dt 2 (1) Using the relation
d 2 x dv dv dx
dv
$
$
$v
2
dt
dt dx dt
dx
we can rewrite (1) as (2) dmax % 2
v0
g "1 ' sin $ # 215.
mg sin θ
θ mg The equation of motion along the plane is
m dv
% mg sin 6 & kmv 2
dt (1) Rewriting this equation in the form
1
dv
% dt
k g sin 6 & v 2
k 42 (2) CHAPTER 2 We know that the velocity of the particle continues to increase with time (i.e., dv dt ! 0 ), so that " g k # sin $ ! v2 . Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We find 1
k 1 v
&
'
tanh %1 *
+(t)C
g
g
*
sin $
sin $ +
k
,k
 (3) The initial condition v(t = 0) = 0 implies C = 0. Therefore,
v( g
sin $ tanh
k " # gk sin $ t ( dx
dt (4) We can integrate this equation to obtain the displacement x as a function of time:
x( g
sin $ . tanh
k " # gk sin $ t dt Using Eq. (E.17a), Appendix E, we obtain
x( " # ln cosh gk sin $ t
g
sin $
) C/
k
gk sin $ (5) The initial condition x(t = 0) = 0 implies C/ = 0. Therefore, the relation between d and t is
d( 1
ln cosh
k " gk sin $ t # (6) From this equation, we can easily find
t( "# cosh %1 e dk gk sin $ (7) 216. The only force which is applied to the article is the component of the gravitational force
along the slope: mg sin 0. So the acceleration is g sin 0. Therefore the velocity and displacement
along the slope for upward motion are described by: v ( v0 % " g sin 0 # t x ( v0 t % 1
" g sin 0 # t 2
2 where the initial conditions v " t ( 0 # ( v0 and x " t ( 0 # ( 0 have been used.
At the highest position the velocity becomes zero, so the time required to reach the highest (1)
(2) "2! gs # "VB2 VB # 2! gs VB # 15.6 m/sec
225.
a) At A, the forces on the ball are:
N mg The track counters the gravitational force and provides centripetal acceleration
N " mg # mv 2 R
Get v by conservation of energy:...
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This note was uploaded on 12/14/2013 for the course PHYS 301 taught by Professor Argryes during the Fall '13 term at University of Cincinnati.
 Fall '13
 Argryes
 Physics, mechanics

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