Solutions to Selected Introductory and Additional Problems: Week 15
SOLUTIONS TO SELECTED INTRODUCTORY PROBLEMS (Ch 14)
5. A 1.1 kg racquet has a moment of inertia about a grip axis of rotation of 0.4 kg(m2). What is
its radius of gyration?
m = 1.1 kg
I = 0.4 kg•m2
k=?
I = mk2
0.4 = 1.1k2
k = 0.60 m
6. How much angular impulse must be supplied by the hamstrings to bring a leg swinging at 8
rad/s to a stop, given that the leg’s moment of inertia is 0.7 kg(m2)?
I = 0.7 kg•m2
Angular impulse = Tt
ω = 8 rad/s,
Tt = ∆H
Tt = (Iω1) – (Iω2)
Tt = (0.7)(8) – 0
Tt = 5.6 kg•m2/s
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7. Given the following principal transverse axis moments of inertia and angular velocities,
calculate the angular momentum of each of the following gymnasts. What body configurations
do these moments of inertia represent?
H = Iω
A.) H = 3.5(20) = 70 kg(m2)/s
B.) H = 7.0(10) = 70 kg(m2)/s
C.) H = 15(4.67) = 70 kg(m2)/s
8. A volleyball player’s 3.7 kg arm moves at an average angular velocity of 15 rad/s during
execution of a spike. If the average moment of inertia of the extending arm is 0.45 kg(m2), what
is the average radius of gyration of the arm during the spike?
m = 3.7 kg, ω = 15 rad/s, I = 0.45 kg(m2), k = ?
I = mk2
k2 = I/m
k = 0.35 m
9. A 50 kg diver in a full layout position, with a total body radius of gyration with respect the
her transverse principal axis equal to 0.45 m, leaves a springboard with an angular velocity of 6
rad/s. What is the diver’s angular velocity when she assumes a tuck position, reducing her radius
of gyration to 0.25 m?
m = 50 kg, kL = 0.45 m, ωL = 6 rad/s, kT = 0.25 m, ωT = ?
HL = HT
conservation of angular momentum
HL = mk2ω
HL = 50(0.452)(6) = 60.75
60.75 = 50(0.252)ωT
ω = 19.44 rad/s
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SOLUTIONS TO SELECTED ADDITIONAL PROBLEMS (Ch 14)
1. The radius of gyration of the thigh with respect to the transverse axis at the hip is 54% of the
segment length. The mass of the thigh is 10.5% of total body mass, and the length of the thigh is
23.2% of total body height. What is the moment of inertia of the thigh with respect to the hip for
males of the following body masses and heights? (A = 0.25 kg.m2,
B = 0.32 kg.m2, C = 0.30 kg.m2, D = 0.37 kg.m2)
Mass(kg)
60
60
70
70
A
B
C
D
Height (m)
1.6
1.8
1.6
1.8
I = mk2
A.
I = (60 kg)(0.105)[(1.6 m)(0.232)(0.54)]2 = 0.25 kg.m2
B.
I = (60 kg)(0.105)[(1.8 m)(0.232)(0.54)]2 = 0.32 kg.m2
C.
I = (70 kg)(0.105)[(1.6 m)(0.232)(0.54)]2 = 0.30 kg.m2
D.
I = (70 kg)(0.105)[(1.8 m)(0.232)(0.54)]2 = 0.37 kg.m2
4. A 7.27 kg shot makes seven complete revolutions during its 2.5 s flight. If its radius of
gyration is 2.54 cm, what is its angular momentum? (0.0825 kg.m2/s)
First, calculate average angular velocity:
ω = (7 rev)(2 π rad/rev) / 2.5 s = 17.6 rad/s
Then calculate angular momentum: H = mk2ω
H = (7.27 kg)(0.0254 m)2(17.6 rad/s)
H = 0.0825 kg.m2/s
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5. What is the resulting angular acceleration of a 1.7 kg forearm and hand when the forearm
flexors, attaching 3 cm from the center of rotation at the elbow, produce 10 N of tension, given a
90° angle at the elbow and a forearm and hand radius of gyration of
20 cm? (4.41 rad/s2)
T = Iα
or
Fd⊥ = mk2α
(10 N)(0.03 m) = (1.7 kg)(0.20 m)2α
α = 4.41 rad/s2
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