hiarker (srh959) – homework 22 – Turner – (58220)
1
This
printout
should
have
11
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A uniform rod of mass 2
.
6 kg is 21 m long. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 21 m from the center of
mass of the rod.
Initially the rod makes an
angle of 51
◦
with the horizontal. The rod is
released from rest at an angle of 51
◦
with the
horizontal, as shown in the figure.
21 m
21 m
2
.
6 kg
O
51
◦
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
m ℓ
2
.
Correct answer: 0
.
818255 rad
/
s.
Explanation:
Let :
ℓ
= 21 m
,
θ
= 51
◦
,
and
m
= 2
.
6 kg
.
Rotational kinetic energy is
K
R
=
1
2
I ω
2
and gravitational kinetic energy is
K
trans
=
m g d .
The inertia of the system is
I
=
I
cm
+
m d
2
=
1
12
m ℓ
2
+
m ℓ
2
=
13
12
m ℓ
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
=
m g ℓ
sin
θ
13
24
m ℓ
2
ω
2
=
m g ℓ
sin
θ
ω
2
=
24
13
g
sin
θ
ℓ
ω
=
radicalbigg
24
g
sin
θ
13
ℓ
=
radicalBigg
24 (9
.
8 m
/
s
2
) sin 51
◦
13 (21 m)
=
0
.
818255 rad
/
s
.
keywords:
002
10.0 points
A circularshaped object of mass 15 kg has
an inner radius of 13 cm and an outer radius
of 30 cm. Three forces (acting perpendicular
to the axis of rotation) of magnitudes 13 N,
27 N, and 13 N act on the object, as shown.
The force of magnitude 27 N acts 31
◦
below
the horizontal.
13 N
13 N
27 N
31
◦
ω
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
hiarker (srh959) – homework 22 – Turner – (58220)
2
Find the magnitude of the net torque on
the wheel about the axle through the center
of the object.
Correct answer: 4
.
29 N
·
m.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 NUI
 Physics, Mass, Work, Moment Of Inertia, Correct Answer, kg, Wnet

Click to edit the document details