279 Chapter 8 8.4 The object is in both translational and rotational equilibrium. Thus, we may write: 0 0 x x xF F RΣ = ⇒ − = 0 0y y y gF F R FΣ = ⇒ + − = and ( ) ( )0 cos sin cos 02O y x gF F Fτ θ θ θ Σ = ⇒ − − = ll l 8.6 Resolve the 100-N force into components parallel to and perpendicular to the rod, as ()()100 N cos 20.0 37.0 54.5 NparallelF= ° + ° = and ()().100 N sin 20.0 37.0 83.9 NperpF= ° + ° = The torque due to the 100-N force is equal to the sum of the torques of its components. Thus, ()()()()54.5 N 0 83.9 N 2.00 m 168 N mτ= − = − ⋅ 20.0°20.0°FparallelPivot100 N37.0°Fperp.2.00 m280 CHAPTER 8 8.17 Consider the torques about an axis perpendicular to the page through the left end of the rod. 3.00 m4.00 m6.00 m500 NTx = Tsin 30.0°Ty = Tcos 30.0°100 NRyRx ()()()()( )100 N 3.00 m 500 N 4.00 m0 6.00 m cos 30.0Tτ+Σ = ⇒ =° 443 NT = ()0 sin 30.0 443 N sin 30.0x xF R TΣ = ⇒ = ° = ° 221 N toward the rightxR = 0 cos 30.0 100 N 500 N 0y yF R TΣ = ⇒ + ° − − = ( )600 N 443 N cos30.0 217 N upwardyR = − ° =Rotational Equilibrium and Rotational Dynamics 281 8.37 The initial angular velocity of the wheel is zero, and the final angular velocity is 50.0 m s40.0 rad s1.25 mfvrω= = = Hence, the angular acceleration is 240.0 rad s 083.3 rad s0.480 sf itω ωα−−= = =∆ The torque acting on the wheel is kf rτ= ⋅, so Iτ α= gives ()()2 23110 kg m 83.3 rad s7.33 10 N1.25 mkIfrα⋅= = = × Thus, the coefficient of friction is 347.33 10 N0.5241.40 10 Nkkfnµ×= = =× 8.40 As the bucket drops, it loses gravitational potential energy. The spool gains rotational kinetic energy and the bucket gains translational kinetic energy. Since the string does not slip on the spool, v rω= where r is the radius of the spool. The moment of inertia of the spool is 212I Mr=, where M is the mass of the spool. Conservation of energy gives ()()t r g t r gf iKE KE PE KE KE PE+ + = + + 2 21 10 02 2f imv I mgy mgyω+ + = + + or ( )( )22 21 1 12 2 2i fm r Mr mg y yω ω + = − This gives ()()( )( )( )( )( )222112222 3.00 kg 9.80 m s 4.00 m10.9 rad s3.00 kg+ 5.00 kg 0.600 mi fmg y ym M rω−= = =+ wImurv282 CHAPTER 8 8.52 The initial angular velocity of the system is 2 radrev0.20 0.40 rad ss 1 reviπω π = = The total moment of inertia is given by ( ) ( )( )22 2 21 180 kg 25 kg 2.0 m2 2man cylinderI I I mr MR r= + = + = + Initially, the man is at 2.0 mr= from the axis, and this gives 2 23.7 10 kg miI= × ⋅. At the end, when 1.0 mr=, the moment of inertia is 2 21.3 10 kg mfI= × ⋅. (a) From conservation of angular momentum, f f i iI Iω ω=, or ( )2 22 23.7 10 kg m0.40 rad s 1.14 rad s 3.6 rad s1.3 10 kg mif ifIIω ω π π × ⋅= = = = × ⋅ (b) The change in kinetic energy is 2 21 12 2f f f iKE I Iω ω∆ = −, or ( ) ( )2 22 2 2 21 rad 1 rad1.3 10 kg m 1.14 3.7 10 kg m 0.40 2 s 2 sKEπ π ∆ = × ⋅ − × ⋅ or 25.4 10 JKE∆ = ×. The difference is the work done by the man as he walks inward. 8.55 (a) From conservation of angular momentum, f f i iI Iω ω=, so 11 2if i ofIII I Iω ω ω = = + (b) ( )22 2 21 1 11 2 11 2 1 2 1 21 1 12 2 2f f f o o iI I IKE I I I I KEI I I I I Iω ω ω = = + = = + + + or 11 2fiKEIKE I I=+. Since this is less than 1.0, kinetic energy was lost.Rotational Equilibrium and Rotational Dynamics 283 8.72 Choose 0gPE= at the level of the base of the ramp. Then, conservation of mechanical energy gives ()()trans rot g trans rot gf iKE KE PE KE KE PE+ + = + + ( )( )( )22 21 10 0 sin 02 2iivmg s mv mRRθ + + = + + or ( )()( )222 2223.0 m 3.0 rad s24 msin sin9.80 m s sin 20iiRvsg gωθ θ= = =
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