A mass m, is connected by a light cord to a mass m, as shown. Cord passes over a pulley of
radius. Rand moment of inertial. The acceleration of two masses is a.
TE
m2
T
Answers
Answer:
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Explanation:
ANSWER
The objects meet when each has travelled distance h
The lost PE=(m
1
−m
2
)gh
This equals the gain in KE, so
KE=(m
1
−m
2
)gh=
2
1
(m
1
+m
2
)v
2
+
2
1
I(
R
V
)
2
Assuming that Rwas given in your diagram as the radius of the pulley.
(m
1
−m
2
)gh=
2
1
(m
1
+m
2
+I/R
2
)v
2
v=
m
1
+m
2
+I/R
2
2(m
1
−m
2
)gh
vα
m
1
+m
2
+
R
2
I
m
1
−m
2
The objects meet when each has travelled distance h
The lost PE=(m
1
−m
2
)gh
This equals the gain in KE, so
KE=(m
1
−m
2
)gh=
2
1
(m
1
+m
2
)v
2
+
2
1
I(
R
V
)
2
Assuming that Rwas given in your diagram as the radius of the pulley.
(m
1
−m
2
)gh=
2
1
(m
1
+m
2
+I/R
2
)v
2
v=
m
1
+m
2
+I/R
2
2(m
1
−m
2
)gh
vα
m
1
+m
2
+
R
2
I
m
1
−m
2
ANSWER
The objects meet when each has travelled distance h
The lost PE=(m
1
−m
2
)gh
This equals the gain in KE, so
KE=(m
1
−m
2
)gh=
2
1
(m
1
+m
2
)v
2
+
2
1
I(
R
V
)
2
Assuming that Rwas given in your diagram as the radius of the pulley.
(m
1
−m
2
)gh=
2
1
(m
1
+m
2
+I/R
2
)v
2
v=
m
1
+m
2
+I/R
2
2(m
1
−m
2
)gh
vα
m
1
+m
2
+
R
2
I
m
1
−m
2
The objects meet when each has travelled distance h
The lost PE=(m
1
−m
2
)gh
This equals the gain in KE, so
KE=(m
1
−m
2
)gh=
2
1
(m
1
+m
2
)v
2
+
2
1
I(
R
V
)
2
Assuming that Rwas given in your diagram as the radius of the pulley.
(m
1
−m
2
)gh=
2
1
(m
1
+m
2
+I/R
2
)v
2
v=
m
1
+m
2
+I/R
2
2(m
1
−m
2
)gh
vα
m
1
+m
2
+
R
2
I
m
1
−m
2