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 L=I 4. A particle is located at r=Oi+3j+Ok, in meter. A constant force F=Oi+0j+4k (in Newton's)begins to act on the particle. As the particle accelerates under the action of this force, the torque as measured ab out the origin (A)increases.(B)decrease (C)is zero. (D)is a nonzero constant Solution: The torque as measured about the origin is z=F×F=(0+3j+0k)×(0i+0j+4k)=12i So it is a nonzero constant I Filling the blank 1. The total angular momentum of the system of i at o --202 (4.4mvo kg m /s)k -4.00 Solution: Using the definition of the angular -200 Vo momentum=F×P=F×m, the total angular momentum of the system of particles is Loa=(-4+2j)×3m(-v0i)+(li+2))×2m(v01)+(1.21-2j×2m(vj) 6mvok-4mvok+2.4my k= 4.4mvok(kg. m2/s) 2. A particle of mass 137g 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 angular momentum of the particle about the origin is_62.472 kg m / s Solution: Using the definition of the angular momentum, the angular momentum of the particle about the L=F×m=msin6=lm=12×10-3×137×10-3×380=625×10-2kg·m2/s 3. The rotor of an electric motor has a rotational inertia Im=2.47x10. m about its central axis The motor is mounted parallel to the axis of a space probe having a rotational inertia Ip=12.6kg. m about its axis. The number of revolutions of the motor required to turn the probe through 25.00 about its axis is 3 54rev Solution: assume the two axes is coaxial. the angular momentum is conserved we have LOm=(m+l)onot 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 , ω r r L = I 4. A particle is located at r i j k ˆ = 0ˆ + 3 ˆ + 0 r , in meter. A constant force F i j k ˆ = 0ˆ + 0 ˆ + 4 r (in 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 （ D ） (A) increases. (B) decreases. (C) is zero. (D) is a nonzero constant. Solution: The torque as measured about the origin is r F i j k i j k i 12ˆ )ˆ ) (0ˆ 0 ˆ 4 ˆ = × = (0ˆ + 3 ˆ + 0 × + + = v v v τ 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 is mv k ˆ (4.4 kg m /s) 2 0 ⋅ Solution: Using the definition of the angular momentum L r P r mv v v v v v = × = × , the total angular momentum of the system of particles is (kg m /s) ˆ 4.4 ˆ 2.4 ˆ 4 ˆ 6 )ˆ ) 2 ( 2ˆ 2 ˆ ) (1. ˆ ) 2 ( 1ˆ 2 ˆ ) ( ˆ ) 3 ( 4ˆ 2 ˆ ( 2 0 0 0 0 0 0 0 = − + = ⋅ = − + × − + + × + − × mv k mv k mv k mv k L i j m v i i j m v i i j m v j total v 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 angular momentum of the particle about the origin is 62.472 kg·m2 /s . Solution: Using the definition of the angular momentum, the angular momentum of the particle about the origin is sin 12 10 13.7 10 380 6.25 10 kg m /s 3 3 2 2 = × = = = × × × × = × ⋅ − − − L r mv rmv θ dmv v v v 3. The rotor of an electric motor has a rotational inertia Im=2.47×10-3kg⋅m 2 about its central axis. The motor is mounted parallel to the axis of a space probe having a rotational inertia Ip=12.6kg⋅m 2 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 is coaxial, the angular momentum is conserved, we have Imω m = (Im + I p )ω y(m) Fig.1 2.00 3m x(m) 2m 2m -4.00 -2.00 1.00 1.20 v i ˆ 0 v i ˆ − 0 v j ˆ 0 O