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 KImv=+122122ω(Eq. 11-5), where I = mR2is 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: ω= v/R. Thus, ()(22222211140 kg0.150 m/s22vKmRmvmvR⎛⎞=+==⎜⎟⎝⎠),which implies that the work required is – 3.15 J. 23. If we write then (using Eq. 3-30) we find Grxyz=++±±±ijkGGrF×is equal to yFzFzFxFxFyFzyxzyx−+−+−dibgdi±±ij±k.(a) Plugging in, we find ()()()()ˆˆ3.0m6.0N4.0m8.0Nk(50N m) k.τ=−−=⎡⎤⎣⎦G⋅(b) We use Eq. 3-27, where φis the angle between ||sin,GGrFrF×=φGrand . Now GFrxy=+=2250.mand FFFxy=+=2210 N. Thus, rF==501050.mNN m,bgbg⋅the same as the magnitude of the vector product calculated in part (a). This implies sin φ= 1 and φ = 90°. 29. (a) We use where GAGG=×mrv,Gris the position vector of the object, Gvis its velocity vector, and mis its mass. Only the xand zcomponents of the position and velocity vectors are nonzero, so Eq. 3-30 leads to GGrvxvzvzz×= −+b±j.gTherefore, ()()()()()()()ˆˆj0.25 kg2.0 m5.0 m s2.0 m5.0 m sj0.zxmxvzv=−+=−+ −−=GA(b) If we write then (using Eq. 3-30) we find Grxyz=++±±±ijk,GGrF×is equal to yFzFzFxFxFyFzyxzyx−+−+−dibgdi±±.ij±kWith x= 2.0, z= –2.0, Fy= 4.0 and all other components zero (and SI units understood) the expression above yields
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