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rote (cmr2495) – homework 22 – Turner – (58120)
1
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beFore answering.
001
10.0 points
A uniForm rod oF mass 3 kg is 15 m long. The
rod is pivoted about a horizontal, Frictionless
pin at the end oF a thin extension (oF negligible
mass) a distance 15 m From the center oF
mass oF the rod. Initially the rod makes an
angle oF 53
◦
with the horizontal. The rod is
released From rest at an angle oF 53
◦
with the
horizontal, as shown in the fgure.
15 m
15 m
3 kg
O
53
◦
What is the angular speed oF the rod at
the instant the rod is in a horizontal position?
The acceleration oF gravity is 9
.
8 m
/
s
2
and the
moment oF inertia oF the rod about its center
oF mass is
I
cm
=
1
12
mℓ
2
.
Correct answer: 0
.
981467 rad
/
s.
Explanation:
Let :
ℓ
= 15 m
,
θ
= 53
◦
,
and
m
= 3 kg
.
Rotational kinetic energy is
K
R
=
1
2
I ω
2
and gravitational kinetic energy is
K
trans
=
mg d .
The inertia oF the system is
I
=
I
cm
+
md
2
=
1
12
mℓ
2
+
mℓ
2
=
13
12
mℓ
2
.
Since the rod is uniForm, its center oF mass
is located a distance
ℓ
From the pivot. The
vertical height oF the center oF mass above
the horizontal is
ℓ
sin
θ .
Using conservation oF
energy,
K
i
+
U
i
=
K
f
+
U
f
K
f
=
U
i
1
2
I ω
2
=
mg ℓ
sin
θ
13
24
mℓ
2
ω
2
=
mg ℓ
sin
θ
ω
2
=
24
13
g
sin
θ
ℓ
ω
=
r
24
g
sin
θ
13
ℓ
=
R
24 (9
.
8 m
/
s
2
) sin 53
◦
13 (15 m)
=
0
.
981467 rad
/
s
.
keywords:
002
10.0 points
A circularshaped object oF mass 11 kg has
an inner radius oF 14 cm and an outer radius
oF 27 cm. Three Forces (acting perpendicular
to the axis oF rotation) oF magnitudes 13 N,
27 N, and 15 N act on the object, as shown.
The Force oF magnitude 27 N acts 36
◦
below
the horizontal.
13 N
15 N
27 N
36
◦
ω
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View Full Documentrote (cmr2495) – homework 22 – Turner – (58120)
2
Find the magnitude of the net torque on
the wheel about the axle through the center
of the object.
Correct answer: 3
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 Spring '10
 McCord

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