大学物理课后习题答案.A1-8

not be constant. (D) Neither the angular momentum nor the angular velocity necessarily has a constant direction. Solution: Using conservation of angular momentum and the definition of the angular momentum ,
Solution: The angular momentum of this particle about the origin is L = r × P , so the position
r
v
v
j. vector of the particle when t = 0s is r = −2 ˆ
Assume the mass of the particle is m, the position vector of the particle at any time t is
r
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ˆ , and when the force acts in the negative x of the torque about the origin is 2.21 ˆ j + 0.936k ˆ. j − 0.936k direction, the components of the torque about the origin is − 2.21 ˆ
y(m)
ˆ − v0 i
3m
2.00 2m 1.20 1.00 O -2.00 Fig.1 2m
ˆ v0i
x(m)
ˆ is ( 4.4mv0 kg ⋅ m /s) k
2
-4.00 the total
Solution: According to the definition of the angular momentum
r
r r L = Iω
r
ˆ) × (0i ˆ) = 12i ˆ+0ˆ ˆ τ = r × F = (0iˆ + 3 ˆ j + 0k j + 4k
v
v
v
So it is a nonzero constant.
II. Filling the Blanks
1. The total angular momentum of the system of particles pictured in Figure 1 about the origin at O
I mωm = ( I m + I p )ω
Integrate both sides of the equation with respect to time, we get
I mθ1 = ( I m + I p )θ 2 ( I + I )θ θ n = 1 = m o p 2 = 354rev 2π 360 I m
momentum of this particle is (A) decreasing (B) constant. (C) increasing. （ (D) possibly but not necessarily constant. B ）
r r 2 ˆ when t = 0s. The magnitude of the angular momentum about the origin of L = (20kg ⋅ m /s) k
4. Two particles each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. The total angular momentum of the system about any origin is mvd . Solution: Select a point O as an origin and set up coordinate system as shown in figure. The total angular momentum of the system about the origin is
v
p ˆ v ˆ−2ˆ r = vti j = ti −2ˆ j m
So the angular momentum of this particle is
r v v p ˆ ˆ = (20kg ⋅ m 2 /s)k ˆ ˆ = 2 pk L = r × P = ( ti −2ˆ j ) × Pi m
ˆ , in meter. A constant force F = 0i ˆ (in ˆ + 3ˆ ˆ+0ˆ 4. A particle is located at r = 0i j + 0k j + 4k
Newton’s) begins to act on the particle. As the particle accelerates under the action of this force，the torque as measured about the origin is (A) increases. (B) decreases. (C) is zero. (D) is a nonzero constant. Solution: The torque as measured about the origin is （ D ）
ˆ+4ˆ 1. A particle moves with position given by r = 3ti j , where
momentum of this particle about the origin is (A) increasing in time. origin is (B) constant in time. is measured in seconds. For each of the following, consider only t > 0. The magnitude of the angular
∫
θ1
0
I mωm dt = ∫ ( I m + I p )ωdt
0
θ2
v v v v v v v Ltotal = L1 + L2 = r1 × mv + r2 × mv r = (r1mv sin θ1 + r2 mv sin θ 2 )k r r = mv(r1 sin θ1 + r2 sin θ 2 )k = mvdk
Thus the magnitude of the angular momentum of this particle L = 20kg ⋅ m /s = constant .
2
3. A solid object is rotating freely without experiencing any external torques. In this case （ A ） (A) Both the angular momentum and angular velocity have constant direction. (B) The direction of angular momentum is constant but the direction of the angular velocity might not be constant. (C) The direction of angular velocity is constant but the direction of the angular momentum might
University Physics AI No. 8 Spin and Orbital Motion
Class Number
v
Name
v r is measured in meters when t
（ B ） (C) decreasing in time. (D) undefined
I. Choose the Correct Answer
Thus the magnitude of the angular momentum of this particle L = 12m kg ⋅ m /s = constant .
2
ˆ . The particle has an angular 2. A particle moves with constant momentum p = (10kg ⋅ m/s)i
0 0 0 0
2. A particle of mass 13.7g is moving with a constant velocity of magnitude 380m/s. The particle, moving in a straight line, passes with the distance 12cm to the origin. The magnitude of the angular momentum of the particle about the origin is 62.472 kg·m2/s . Solution: By the definition of the angular momentum, the angular momentum of the particle about the origin is
ˆ 2 − v0 i
y 1 r1
ˆ v0i
d x
r2
O θ2
θ1
ˆ . A constant force of ˆ + (−0.36m) ˆ j + (0.85m)k 5. A particle is located at r = (0.54m)i
magnitude 2.6 N acts on the particle. When the force acts in the positive x direction, the components
v v v L = r × mv = rmv sin θ = dmv = 12 × 10−3 × 13.7 × 10−3 × 380 = 6.25 × 10−2 kg ⋅ m 2 /s
3. The rotor of an electric motor has a rotational inertia Im=2.47×10-3kg⋅m2 about its central axis. The motor is mounted parallel to the axis of a space probe having a rotational inertia Ip=12.6kg⋅m2 about its axis. The number of revolutions of the motor required to turn the probe through 25.0° about its axis is 354rev . Solution: Assume the two axes are coaxial, the angular momentum is conserved, we have
Solution: Assume the mass of the particle is m, so the angular momentum of this particle about the
v r v v v dr v v ˆ ˆ+4ˆ ˆ) = −12mk L = r × P = r × mv = r × m j ) × m(3i = (3ti dt
v v v v v L = r × P = r × mv ,
v0 ˆ j
angular momentum of the system of particles is
v ˆ+2ˆ ˆ) + (1i ˆ+2ˆ ˆ) + (1.2i ˆ−2ˆ Ltotal = (−4i j ) × 3m(−v0 i j ) × 2 m (v 0 i j ) × 2m(v 0 ˆ j) ˆ − 4mv k ˆ + 2.4mv k ˆ = 4.4mv k ˆ(kg ⋅ m 2 /s) = 6mv k