lew (dl9564) – hk3 – Opyrchal – (41104)
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001
10.0 points
A circular disk with moment of inertia
I
=
1
2
m R
2
mass
m
and radius
R
is mounted at
its center, about which it can rotate freely.
A light cord wrapped around it supports a
weight
m g
.
R
I
ω
m
T
g
Find the total kinetic energy of the system
when the weight is moving at a speed
v
.
1.
K
=
5
4
m v
2
2.
K
=
2
3
m v
2
3.
K
=
1
2
m v
2
4.
K
=
1
3
m v
2
5.
K
=
3
2
m v
2
6.
K
=
m v
2
7.
K
=
3
4
m v
2
correct
8.
K
=
5
2
m v
2
9.
K
=
4
5
m v
2
Explanation:
v
=
r ω ,
so the total kinetic energy is
K
tot
=
K
m
+
K
rot
=
1
2
m v
2
+
1
2
I ω
2
=
1
2
m v
2
+
1
2
parenleftbigg
1
2
m R
2
parenrightbigg
parenleftBig
v
R
parenrightBig
2
=
3
4
m v
2
.
keywords:
002
10.0 points
A 1.5 kg bicycle tire of radius 0.33 m starts
from rest and rolls down from the top of a hill
that is 15.0 m high.
What is the translational speed of the tire
when it reaches the bottom of the hill?
As
sume that the tire is a hoop with
I
=
mr
2
.
The acceleration of gravity is 9
.
81 m
/
s
2
.
Correct answer: 12
.
1305 m
/
s.
Explanation:
Let :
m
= 1
.
5 kg
,
r
= 0
.
33 m
,
h
= 15
.
0 m
,
and
g
= 9
.
81 m
/
s
2
.
The moment of inertia of the solid cylinder
is
I
=
1
2
m r
2
v
i
= 0 m/s and
h
f
= 0 m, so by
conservation of energy,
U
i
=
K
f
U
i
=
K
trans,f
+
K
rot,f
m g h
=
1
2
m v
2
f
+
1
2
I ω
2
f
m g h
=
1
2
m v
2
f
+
1
2
m r
2
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 Fall '08
 opyrachal
 Inertia, Mass, Moment Of Inertia, Rotation, kg

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