Unformatted text preview: Etop # Ttop $ Utop # 0 $ mgh
EA # TA $ U A # 1
mv 2 $ 0
2 Etop # EA % v # 2 gh
So
N # mg $ m2 gh R
2h '
&
N # mg ( 1 $ )
*
R+
b) At B the forces are:
N mg 45˚ N # mv 2 R $ mg cos 45,
# mv 2 R $ mg 2 (1) Get v by conservation of energy. From a), Etotal # mgh .
At B, E # 1
m v 2 $ mgh 2 2 R R 45˚ R R cos 45˚ = R 2 h′ R! R
" h#
2 or 1%
$
h# ! R ' 1 &
(
)
2* So Etotal ! TB " UB becomes:
1% 1
$
2
mgh ! mgR ' 1 &
( " mv
)
2* 2
Solving for v 2
+
1 %,
$
2
2  gh & gR ' 1 &
( !v
)
2*.
/
0
Substituting into (1):
+ 2h $ 3
%,
N ! mg  " '
& 2( .
*0
/R ) 2
2
c) From b) vB ! 2 g + h & R " R
/ 2,
0 1 v ! +2 g h & R " R
/ 2 2,
0 12 d) This is a projectile motion problem
45˚
B 45˚ A Put the origin at A.
The equations:
x ! x0 " vx 0 t
y ! y0 " v y 0 t & 12
gt
2 become
x! R
v
" Bt
2
2 y ! h# "
Solve (3) for t when y = 0 (ball lands). vB
1
t & gt 2
2
2 (2)
(3) gt 2 ! 2 vBt ! 2h # " 0
t" 2
2 vB $ 2vB % 8 gh # 2g We discard the negative root since it gives a negative time. Substituting into (2):
& 2 v $ 2v 2 % 8 g h # '
71
B
B
(
)
2g
2
2(
)
*
+
The equilibrium point (where dU d! " 0 ) that we wish to look at is clearly ! = 0. At that point,
Using the previous expressions for vB and h# yields
we have d 2U d! 2 " mg # R % b 2$ , which is stable for R & b 2 and unstable for R ' b( 2 . We can
12
use the results of Problem 246"to obtainR % h % & hfor 3 R2case 2R " ' 2 , where we will find that
b
x
2 ! 1 stability 2 ! the %
R2 )
(
2
+
the first nontrivial result is in fourth order and*is negative. We therefore have an equilibrium at
! = 0 which is stable for R & b 2 and unstable for R ) b 2 .
e) U ( x) " mgy( x ) , with y(0) " h , so U ( x) has the shape of the track.
NEWTONIAN MECHANICS—SINGLE PAR
RTICLE
vB x" % ,  243.
F " % kx + kx 3 * 2
226. All of the kinetic energy of the block goes into compressing the spring, so that
mv 2 2 " kx 2 2 , or x " v m k...
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Full Document
 Fall '13
 Argryes
 Physics, mechanics, Force, Sin, Trigraph, dt

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