Chapter 11 Student Solutions Manual

Chapter 11 Student Solutions Manual - Chapter 11 Student...

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Chapter 11 – Student Solutions Manual 5. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic energy of the hoop. The initial kinetic energy is KI m v =+ 1 2 2 1 2 2 ω (Eq. 11-5), where I = mR 2 is its rotational inertia about the center of mass, m = 140 kg, and v = 0.150 m/s is the speed of its center of mass. Eq. 11-2 relates the angular speed to the speed of the center of mass: = v / R . Thus, () ( 2 2 22 2 2 11 140 kg 0.150 m/s v Km R m v m v R ⎛⎞ = = ⎜⎟ ⎝⎠ ) , which implies that the work required is – 3.15 J. 23. If we write then (using Eq. 3-30) we find G rxyz =++ ±±± ij k G G r F × is equal to yF zF zF xF xF yF zy xz y x −+ di bg ±± ± k . (a) Plugging in, we find ( ) ( )( ) ˆˆ 3.0m 6.0N 4.0m 8.0N k (50N m) k. τ =− = ⎡⎤ ⎣⎦ G (b) We use Eq. 3-27, where φ is the angle between | | sin , G G rFr F ×= G r and . Now G F rx y = 50 .m and FF F xy = 10 N. Thus, rF == 10 50 .m N Nm , b gb g the same as the magnitude of the vector product calculated in part (a). This implies sin = 1 and φ = 90°. 29. (a) We use where G A GG mr v , G r is the position vector of the object, G v is its velocity vector, and m is its mass. Only the x and z components of the position and velocity vectors are nonzero, so Eq. 3-30 leads to G G rv x v z v zz ×=− + b ± j. g Therefore, ( ) ( ) ( ) j 0.25 kg 2.0 m 5.0 m s 2.0 m 5.0 m s j 0. zx mx v =−+ = + = G A (b) If we write then (using Eq. 3-30) we find G rx yz k , G G r F × is equal to yF zF zF xF xF yF b g . ± k With x = 2.0, z = –2.0, F y = 4.0 and all other components zero (and SI units understood) the expression above yields
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G G G τ =× = + rF 80 . ± . ± ik N e j m . 33. If we write (for the general case) G rx yz =++ ±±± ij k , then (using Eq. 3-30) we find G G r v × is equal to yv zv zv xv xv yv zy xz y x −+ di bg ±± . ijk ± (a) The angular momentum is given by the vector product G A G G mr v , where G r is the position vector of the particle, G v is its velocity, and m = 3.0 kg is its mass. Substituting (with SI units understood) x = 3, y = 8, z = 0, v x = 5, v y = –6 and v z = 0 into the above expression, we obtain () 22 ˆˆ 3.0 [(3.0)( 6.0) (8.0)(5.0)]k ( 1.7 10 kg m s)k. =− −= × G A (b) The torque is given by Eq. 11-14, G G G . We write G rx y =+ ± i ± j k and and obtain G FF x = ± i G =+× = xy F y F xx ± ± i e j e j since and Thus, we find ii ×= 0 ±± ± ji k . ×=− ( ) 8.0m 7.0N k (56N m)k. = G (c) According to Newton’s second law G G A = dd t , so the rate of change of the angular momentum is 56 kg m 2 /s 2 , in the positive z direction. 37. (a) Since τ = dL / dt , the average torque acting during any interval Δ t is given by avg LL t fi Δ , where L i is the initial angular momentum and L f is the final angular momentum. Thus, avg 0.800 kg m s 3.00 kg m s 1.47 N m 1.50s ⋅− == , or avg | | 1.47 N m =⋅ . In this case the negative sign indicates that the direction of the torque is opposite the direction of the initial angular momentum, implicitly taken to be positive.
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Chapter 11 Student Solutions Manual - Chapter 11 Student...

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