09prob_11

09prob_11 - Assigment#9 Page 1 of 4 UC Davis Physics 9A-C Assignment#9

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Unformatted text preview: Assigment #9, 5/27/11 Page 1 of 4 UC Davis Physics 9A-C Assignment #9 Cole [9.82]
a)
No
nonconservative
forces
do
work
so
conserve
energy.

 Eø
=
E

õ

 

 mgh = 1 2 mv 2 + 1 2 Â ω 2 
 where
∑
=
 v/r.

 Solve
for
Â:
 

 Â= 2 gh − v 2 v 2 mr 2 = 2 9.8 
m/s 2 ( ) 2 
m ( ) − 5 
m/s ( ) 2 5 
m/s ( ) 2 8 
kg ( ) 0.37 
m ( ) 2 
=
0.622
kg·m 2 . b)

Moving
mass
toward
the
axis
of
rotation
would
decrease
the
moment
of
inertia.

The
best
we
 can
do
is
move
approximately
all
of
the
mass
to
the
radius.

The
mass
of
the
wheel
is

 

 M = w g = 280 
N 9.8
m/s 2 
=

28.6
kg.

Doing
this
gives

Â
=
 Mr 2 
=
(28.6
kg)(0.37
m) 2 
=
3.92
kg·m 2 .

 Anything
larger
is
impossible. [9.83]

a)

Gravity
acts
at
the
center
of
mass.

The
work
done
by
gravity
is
-
Î U cm 
.
The
potential
 energy
of
the
meter
stick
will
be
least
when
it
is
vertical,
at
which
point,
the
center
of
mass
will
 have
lowered
by
0.5
m

õ

 Î U
=
0
-

mgh

=- (0.16
kg)(9.8
m/s 2 )(0.5
m)
=
-0.784
J


 b)
The
change
in
potential
energy
gets
converted
into
kinetic
energy

Î K
+ÎU
=
0

õ 

 1 2 
Â p 
∑ 2 
=
-
ÎU

õ

 

 ω = − 2 Δ U Â p = 2 − mgh ( ) 1 3 mL 2 = 5.42 rad/s 

where
we
have
used
the
moment
 of
inertia
about
the
pivot. c)

 v
=
r∑

=
L∑

=
 (1
m)(5.42/s)

=
5.42
m/s

d)

 v = 2 gh = 2 9.8 m/s 2 ( ) 1 m ( ) 
=
4.43
m/s.


õ
 v ( c ) = 3/2 v ( d ) ,

that
is,
the
average
linear
acceleration
of
the
end
of
the
stick
is
larger
than
 g !
 [9.85]

The
energies
we
must
account
for
are
the
work
done
by
friction,
KE
of
the
2
masses,
 potential
energy
of
mass
 B ,
and
the
rotational
energy
of
the
pulley.

Take
the
zero
point
for
 potential
energy
to
be
the
starting
height
for
block
(B).

As
(B)
falls,
its
potential
energy
will
 decrease.

Recall,
work
done
by
friction
changes
the
energy
of
the
system:

 W f 
=
E
-
Eø

õ

-μ k
 m A gd
=
( 1 2 m A v 2 + 1 2 m B v 2 +Â
∑ 2-m B gd)
-
0
 

We
need
one
more
relation
connecting
the
 angular
speed
of
the
pulley.

If
the
string
does
not
slip

then

 v
=
R∑.

õ

 

 m B gd − μ m A gd = 1 2 m A + m B + Â R 2 ⎛ ⎝ ⎞ ⎠ v 2 .

Solving
for
 v :


 

 v = 2 m B gd − μ m A gd ( ) m A + m B + Â R 2 ⎛ ⎝ ⎞ ⎠ ....
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This note was uploaded on 09/28/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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09prob_11 - Assigment#9 Page 1 of 4 UC Davis Physics 9A-C Assignment#9

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