12.71. Model:
Assume the string does not slip on the pulley.
Visualize:
The freebody diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward
force on the block
2
,
m
but a downward force on the outer edge of the pulley. Similarly the string exerts a force
on block
1
m
to the right, but a leftward force on the outer edge of the pulley.
Solve:
(a)
Newton’s second law for
1
m
and
2
m
is
11
Tm
a
=
and
22
2
.
gm
a
−
=
Using the constraint
21
,
aa
a
−=
+=
we have
1
a
=
and
.
a
−+
=
Adding these equations, we get
2
()
,
mg
m m a
=+
or
2
1
12
mmg
aT
m
a
mm
=⇒
=
=
++
(b)
When the pulley has mass
m
, the tensions
(
a
n
d
)
TT
in the upper and lower portions of the string are
different. Newton’s second law for
1
m
and the pulley are:
1
2
and
a
T
RT
R I
α
=
−
We are using the minus sign with
because the pulley accelerates clockwise. Also,
.
aR
=
Thus,
a
=
and
2
Ia a
I
R
RR
=
Adding these two equations gives
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.
 Fall '09
 geller
 Force

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