M6 Week 11 Problem Set #11
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M6 Week 11 Problem Set #11
Hallstrom
Problems: 2, 12, 24, 40, 42, 64
2. The gardening tool is used to pull weeds. If a 1.23-N * m torque is required to pull a given
weed, what force did the weed exert on the tool?
τ
=
r
∗
F
F
=
τ
r
F
=
1.23
N
∗
m
0.04
m
F
=
30.75
N →
30.7
N
12.
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0
revolutions. If the CD has a radius of 6.0 cm and a mass of 17 g, what is the torque exerted on
it?
M
=
1
2
mr
2
17
∗
10
−
3
kg
¿
(
6
∗
10
−
2
m
)
2
M
=
1
2
¿
M
=
3.06
∗
10
−
5
ω
2
=
ω
0
2
+
2
αθ
(
450
rev
/
min
¿
2
π
60
rad
/
s
1
rev
/
min
)
2
=
0
2
+
2
α
(
3.0
rev
∗
2
π rad
1
rev
)
α
=
58.9
rad
/
s
2
τ
=
Ma
τ
=
3.06
∗
10
−
5
kg
∗
m
2
∗
58.9
rad
/
s
2
τ
=
1.8
∗
10
−
3
N
∗
m
M6 Week 11 Problem Set #11
Hallstrom
24.
To determine the location of her center of mass, a physics student lies on a
lightweight plank
supported by two scales 2.50 m apart. If the left scale reads 435 N, and the right scale reads 183
N, find
(a)
the student’s mass and
(b)
the distance from the student’s head to her center of
mass.
a.
The student’s mass
∑
f
y
=
f
1
+
f
2
−
mg
=
0
m
=
f
1
+
f
2
g
m
=
435
N
+
183
N
9.8
m
/
s
2
m
=
63.06
kg→
63.1
kg
b.
The distance from the student’s head to her center of mass
∑
τ
¿
=
x
2
f
2
−
x
cg
(
mg
)
=
0
x
cg
=
x
2
f
2
mg
x
cg
=
(
2.5
m
)(
183
N
)
(
63.06
kg
)(
9.8
m
/
s
2
)
x
cg
=
457.5
617.98
x
cg
=
0.74
m
40.
A 2.85-kg bucket is attached to a rope wrapped around a disk-shaped pulley of radius 0.121
m and mass 0.742 kg. If the bucket is allowed to fall,
(a)
what is its linear acceleration?
(b)
What
is the angular acceleration of the pulley?
(c)
How far does the bucket drop in 1.50 s?
(group)
a.
What is its linear acceleration?
For this equation: m1=bucket, m2=mass of pulley
a
=
m
1
g
1
2
m
2
+
m
1
a
=
(
2.85
kg
)(
9.8
m
/
s
2
)
1
2
(
0.742
kg
)+(
2.85
kg
)
a
=
8.68
m
/
s
2
However, this value will be negative because the bucket is falling.
a
=−
8.68
m
/
s
2
b.
What is the angular acceleration of the pulley?
For the purpose of this problem, we are going to use the magnitude of the acceleration,
so the positive value
α
=
a
r
M6 Week 11 Problem Set #11
Hallstrom
α
=
8.68
m
/
s
2
0.121
m
α
=
71.735
rad
/
s
2
→
71.7
rad
/
s
2
c.
How far does the bucked drop in 1.50s?
Again we will use magnitude of the acceleration for this problem. We will also define x
as the distance of the bucket
x
=
1
2
at
2
x
=
1
2
(
8.68
m
/
s
2
)
(
1.5
s
)
2
x
=
9.765
m→
9.76
m
42.
You pull downward with a force of 28 N on a rope that passes over a disk-shaped pulley of
mass 1.2 kg and radius 0.075 m. The other end of the rope is attached to a 0.67-kg mass.
(a)
Is
the tension in the rope the same on both sides of the pulley? If not, which side has the greater
tension?
(b)
Find the tension in the rope on both sides of the pulley.
(group)
a.
Is the tension in the rope the same on both sides of the pulley? If not, which side has the
greater tension?
The tension is not the same on both sides of the pulley. This can be seen by the
calculations below, but the side that the force is applied to will be pulled downward and
have more tension.
b.
Find the tension in the rope on both sides of the pulley.
T
2
−
T
1
=
1
2
m
p
a
T
2
−
m
w
a
−
m
w
g
=
1
2
m
p
a
28
−
(
0.67
)
a
−
(
0.67
) (
9.8
)
=
1
2
(
1.2
)
a
21.43
=
1.27
a
a
=
16.87
m
/
s
2
→
17
m
/
s
2
This acceleration will allow us to solve for
T
1
T
1
=
m
w
a
−
m
w
g
T
1
=
(
0.67
) (
16.87
)
+(
0.67
)(
9.8
)
T
1
=
11.3
+
6.57
T
1
=
17.87
N →
18
N
T
1
<
T
2
18
N
<
28
N
64. A popular make of dental drill can operate at a speed of 42,500 rpm while producing a
torque of 3.68 oz ·in. What is the power output of this drill? Give your answer in watts.
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M6 Week 11 Problem Set #11
Hallstrom
Convert angular speed from rpm to rad/s
ω
=
42500
rpm
∗
(
2
π
60
)
=
4448.33
rad
s
Convert torque from oz*in to N*m
τ
=
3.68
oz
∗
¿
∗
0.007062
=
0.026
N
∗
m
P
=
ω
∗
τ
P
=
0.026
N
∗
m
∗
4448.33
rad
s
P
=
115.656
W →
116
W
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