Integrate both sides of the equation with respect to time, we get LO dt=l(I +I ad m=(m+ln)02 e(Im+1,e2 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 md Solution See figure, select any point O and set up coordinate stem. So the total angular momentum of the system about the origin is tol =l+l,=F xmv+F, xmv = rmvsin 8,+r,mosin 82 =mv(r sin 8,+r, sin 82)=md 5. A particle is located at r=(0.54m)i+(-036m)j+(0.85m)k. A constant force of magnitude 2.6 N acts on the particle. When the force acts in the positive x direction, the components of the torque about the origin is 2.21j+0.936k, and when the force acts in the negative x direction, the components of the torque about the origin is -2. 21j-0.936k Solution: According to the definition of the torque t=Px F when the force is 2.6Ni, the torque about the origin is z=F×F=(0.54-0.36j+0.85k)×(26i)=2.21j+0.936 when the force is -2.6Ni, the torque about the origin is z=F×F=(0.54-0.36j+0.85k)×(-261)=-2.21j-0.936k 6. A cylinder having a mass of l%2kg rotates about its axis of symmetry. F. Forces are applied as shown in Fig. 2: F1 5.88N, F2=4. 13N, and F3=2. 12N. Also R =4.93cm and R2=11.8cm. The magnitude and direction R of the angular acceleration of the cylinder are_ 90.brad/s R↓F30 Solution: Using the definition of the torque t rx F, suppose the clockwise is the positive. The magnitude of the total torque about its axis [=FR-FR2-F,R 588×118×10-2-4.13×118×10-2-212×493×10-2=102N.mIntegrate both sides of the equation with respect to time, we get 354rev 360 ( ) 2 ( ) d ( ) d 1 2 1 2 0 0 1 2 = + = = = + = + ∫ ∫ m m p m m p m m m p I I I n I I I I t I I t o θ π θ θ θ ω ω θ θ 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: See figure, select any point O and set up coordinate system. So the total angular momentum of the system about the origin is mv r r mvd r mv r mv L L L r mv r mv total = + = = + = + = × + × ( sin sin ) sin sin 1 1 2 2 1 1 2 2 1 2 1 2 θ θ θ θ v v v v v v v 5. A particle is located at r i j k ˆ (0.85m) ˆ ( 0.36m) ˆ = (0.54m) + − + r . A constant force of magnitude 2.6 N acts on the particle. When the force acts in the positive x direction, the components of the torque about the origin is j k ˆ 2.21ˆ + 0.936 , and when the force acts in the negative x direction, the components of the torque about the origin is j k ˆ − 2.21ˆ − 0.936 . Solution: According to the definition of the torque r F v v v τ = × , when the force is i 6Nˆ 2. , the torque about the origin is r F i j k i j k ˆ ) (2.6ˆ) 2.21ˆ 0.936 ˆ = × = (0.54ˆ − 0.36 ˆ + 0.85 × = + r v v τ when the force is i 6Nˆ − 2. , the torque about the origin is r F i j k i j k ˆ ) ( 2.6ˆ) 2.21ˆ 0.936 ˆ = × = (0.54ˆ − 0.36 ˆ + 0.85 × − = − − r v v τ 6. A cylinder having a mass of 1.92kg rotates about its axis of symmetry. Forces are applied as shown in Fig.2: F1=5.88N, F2=4.13N, and F3=2.12N. Also R1=4.93cm and R2=11.8cm. The magnitude and direction of the angular acceleration of the cylinder are 90.3rad/s2 . Solution: Using the definition of the torque r F v v v τ = × , suppose the clockwise is the positive. The magnitude of the total torque about its axis of symmetry is 5.88 11.8 10 4.13 11.8 10 2.12 4.93 10 10.2 N m 2 2 2 1 2 2 2 3 1 = × × − × × − × × = ⋅ = − − − − − τ F R F R F R y x v i ˆ 0 v i ˆ − 0 O d 1 2 r1 r2 θ1 θ 2 R1 R2 F3 r F1 r F2 r o 30 Fig.2