Problem #1
An object has an inertial moment of 3 kg m2 and rotates on a fixed axis with an angular velocity of 2 rad/s. Momentum of the object's angle is. . . .?Answer:
It is known: the moment of inertia of an object, I = 3 kgm2, and the angular velocity of an object, ω = 2 rad/s, then the angular momentum of the object is
L = Iω = (3)(2) = 6 kg.m2/s
Problem #2
A particle of 2.0 kg mass rotates at an angle of 4.0 rad / s. The radius of the particle's path is 1.0 meter. The angular momentum of these particles is. . . .?
Answer:
Particle mass, m = 2.0 kg; angular velocity of particle, ω = 4.0 rad/s; radius of particle path, r = 1.0 m,
The moment of inertia of the particle is
I = mr² = (2.0 kg) (1.0 m)2 = 2 kgm2
Then the particle's angular momentum is
L = Iω
L = (2 kgm2) (4 rad / s) = 8 kgm²/s
Angular momentun of particle is 8 kg m²/s
Problem #3
The figure on the side shows the momentary position of objects P and Q which are rotating at a constant speed of 1 m/s and 3 m/s against the O point in the opposite direction.
The total angular momentum at point O is
LO = LP + LQ
LO = –mpvprp + mQvQrQ = – 3 kg x 1 m/s x 5 m + 2 kg x 3 m/s x 4 m
LO = 9 kgm2/s
Problem #4
Calculate the angular momentum for the following particles. Find the angle between the position and the momentum vectors.
(a) r = (4, ─5,3) and p = (1, 4, ─2)
(b) r = (1, ─2,3) and p = (7, ─1,1)
(c) r = (0, 2, 0) and p = (1, 0, 0)
Answer:
The angular momentum can be found either by evaluating the determinant or by using
L = rpsinφ.
We will use the first method to find L. We can find the angle between the momentum and position vectors using
φ = sin─1(L/rp).
We find the magnitudeof the vectors using the 3D form of the Pythagorean Theorem.
(a) r = (4, 5,3) and p = (1, 4, 2)
= ─2i + 11j + 21k
L = [(─2)2 + (11)2 + (21)2]½ = 23.791 kgm2/s
r = [(4)2 + (─5)2 + (3)2]½ = 7.0711 m
p = [(1)2 + (4)2 + (─2)2]½ = 4.5826 kgm/s
φ = sin─1(L/rp) = sin─1[23.791/(7.0711 × 4.5826)] = 47.20
(b) r = (1, ─2,3) and p = (7, ─1,1)
L = i[(─2)(1) ─ (─1)(3)] ─ j[(1)(1) ─ (7)(3)] + k[(1)(─1) ─ (7)(─2)]
= i + 20j + 13k
L = [(1)2 + (20)2 + (13)2]½ = 23.875 kgm2/s
r = [(4)2 + (─5)2 + (3)2]½ = 3.7417 m
p = [(1)2 + (4)2 + (─2)2]½ = 7.1414 kgm/s
φ = sin─1(L/rp) = sin─1[23.875/(3.7417 × 7.1414)] = 63.30
(c) r = (0, 2, 0) and p = (1, 0, 0)
L = i[(2)(0) ─ (0)(0)] ─ j[(0)(0) ─ (1)(0)] + k[(0)(0) ─ (1)(2)] = ─2k
L = [(0)2 + (0)2 + (─2)2]½ = 2 kgm2/s
r = [(0)2 + (2)2 + (0)2]½ = 2 m
p = [(1)2 + (0)2 + (0)2]½ = 1 kgm/s
φ = sin─1(L/rp) = sin─1[2/(2 x 1)] = 900
Problem #5
Calculate the angular momentum of a phonograph record (LP) rotating at 100/3 rev/min. An LP has a
radius of 15 cm and a mass of 150 g. A typical phonograph can accelerate an LP from rest to its final
speed in 0.35 s, what average torque would be exerted on the LP?
Answer:
The angular momentum of a rotating body is
L = Iω.
An LP is a solid disk. Consulting a table of moments of inertia, we find I = ½MR2. The angular velocity must be converted to rad/s
ω = 100/3 rev/min × 2π rad / rev × 1 min / 60 s = 3.4907 rad/s
Thus we find the angular momentum of the LP to be
L = Iω = ½MR2ω = ½(0.15 kg)(0.15 m)2(3.4907 rad/s) = 5.8905 × 10─3 kgm2/s .
Torque is equal to the change in angular momentum with time
τ = ΔL/Δt = (Lf ─ Li)/Δt = (5.8905 × 10─3 kgm2/s ─0)/0.35 s = 1.68 × 10─2 N.m
Problem #6
Determine the magnitude LO of the angular momentum of the 2.0 kg sphere about point O by using the vector definition of angular momentum and. The center of the sphere lies in the x-y plane.
Vector position of the ball given by
r = (4.0i + 3.0j) m
Vector ball velocity given by
v = vxi + vyj = (8.0 cos 530)i + (8.0 sin 530)j
Vector linear momentum ball is given by
p = mv = (16.0 cos 530)i + (16.0 sin 530)j kg.m/s
then the vector of angular momentum of the ball is obtained from
L = r x p = (4.0i + 3.0j) x [(16.0 cos 530)i + (16.0 sin 530)j]
L = 51.2k ─ 28.8k = 22.4k kg.m/s
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