1. Show that if the angular momentum of a boby is determined with respect to an arbitrary point A, then. r r r. H r A can be expressed by H r r r r

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1 1. Shw that if the angula entu f a bb is deteined with espect t an abita pint, then H can be epessed b H = ρ / v + H. This equies substituting ρ = ρ + ρ / int H = ρ d v + ρ ( ω ρ ) d and epanding, nte that ρd = 0 b definitin f the ass cente and v = v + ω ρ /. Sl) inseting ρ H Then, = ρ + ρ int Eq. H = ρ d v + ρ ( ω ρ ) d / = ( ρ + ρ / ) d v + ( ρ + ρ / ) [ ω ( ρ + ρ / )] d = ρ d v + ρ / d v + ρ ( ω ρ ) d + ρd ( ω ρ / ) + ρ ω ρd) + d[ ρ / ( ω ρ H / ( / (because ρd = 0 & ρ / and ω : Independent t the integatin.) = ρ / v (because + H + ρ ( ω ρ )] [ / / d = : Ttal ass and H = ρ ( ω ρ ) d [ v + ( ω ρ / = ρ / )] + H [F 3D kineatics, v = + ω ρ ) ] )] v ( / Finall, H ρ = / v + H (NSWER)

2 . Rd has a weight f 6 lb and is attached t tw sth cllas at its end pints b ball-and-scket jints. If clla is ving dwnwad at a speed f 8 ft/s, deteine the kinetic eneg f the end at the instant shwn. ssue that at this instant the angula velcit f the d is diected pependicula t the d s ais. Sl) Let s suppse : Pincipal aes Ments f inetia 1 6 I ' = I ' = ( )(7) = 0.76 slug ft & I ' = Since ω, let s assue ω = ωî ' (alng ais) Then, H = H ngula velcit ' iˆ' = I ωiˆ ' ' ( H & H 0 ) ' ' = In de t deteine the angula velcit, let s cnside the fied efeences. F 3D kineatics f a igid bd, = + ω whee = ( 8) kˆ /s, v v v / v ω = ω iˆ + ω ˆj + ω kˆ = ˆ i + 6 ˆj / 3kˆ = v iˆ /s v iˆ iˆ ˆj kˆ 8kˆ + ω ω ω = 6 3 Then, 3 ω 6ω = v + ω 3 ω = 0 6 ω ω 8 = 0 (4 unknwns, but 3 equatins)

3 Since ω, ω = ω + 6ω 3ω = 0 (4 th equatin) ω = ad/s, ω = ad/s, ω = ad/s, v = 1. 0 ft/s = Then, ω ω + ω + ω = (0.98) + ( 1.06) + ( 1.47) = 4. 4 ad/s Kinetic eneg, T = v + ω H = v + I ' ω (, I, ω : Knwn, but v?) T deteine v, v = v iˆ ˆj kˆ 8 ˆ 1 + ω / = k (because / = / ) = 6.0ˆ i 4.0kˆ ft/s Finall, T = v + I ' ω = ( )[(6.0) + ( 4.0) ] v + (0.76)(4.4) = 6.46 ft lb 3.

4 3. The -kg thin disk is cnnected t the slende d which is fied t the ball-and-scket jint at. If it is eleased f est in the psitin shwn (psitin C), deteine the spin f the disk abut the d when the disk eaches its lwest psitin (psitin D). Neglect the ass f the d and the disk lls withut slipping. (Hint: The kinetic eneg f D, the disk at the lwest psitin D is equivalent t the gavitatinal ptential eneg due t the height diffeence between C and D.) Sl) Let s suppse : Pincipal aes Ments f inetia 1 I ' = I ' = ()(0.1) + ()(0.5) = kg 4 (Paallel-ais thee) 1 I ' = ()(0.1) = 0.01 kg h D θ R 30, h 1 ngula velcit ω = ω ' ˆ' j (Spin f the disk) + ω kˆ (Rtatin abut ais) whee ˆ sin ˆ' cs ˆ 0.1 ω k = ω θ j + ω θ k' ( tan θ = 0.5 = ( 0.0ω ω ') ˆ' j ω kˆ' θ = 11.3 ) disk t the pint D, disk ( ω ' ) = R( ω ) ω = ω' = sinθω' = 0. 0ω ' R Then, ω = 0.96ω ˆ' 0.0 ˆ ' j + ω k' Since Kinetic eneg, T at D: g h 1 + h ) (Eneg pinciple) ( 1 T = ( I ' ω ' + I ' ω' + I ' ω' ) = g( h1 + h ) (Oigin : a fied pint) 1 = = whee h = (0.5)sin( ) & h = (0.5)sin( ) = [(0.01)( 0.96ω ' ) + (0.505)(0.0ω ') ] = ()(9.81)(0.49) ω (Spin f the disk) = 6. ad/s (NSWER) '

5 4. thin unif plate having a ass f 0.4 kg is spinning with a cnstant angula velcit ω = 100 ad/s abut its diagnal as shwn in figue. If the pesn hlding the cne f the plate at eleases his finge, the plate will fall dwnwad n its side C. T pevent this f happening, we have t appl the necessa cuple ent M. Since the given aes,, ae the pincipal aes, we can use the Eule s equatins f tin t find the ent M. using the pincipal ents f inetia, I, I, and I and ω, ω, and ω, please deteent the cpnents f ent M, M, and M. Sl) Using the pincipal aes f efeence shwn ( ), 1 3 I = (0.4)(0.3) = 3 10 kg I = (0.4)(0.15) = kg I = (0.4)[(0.3) + (0.15) ] = kg 1 and the angula velcit and angula acceleatin f the plate ω = ω sin θ = 100sin 6.6 & ω& = 0 (because ω = ω cs θ = 100cs 6.6 & ω& = 0 ω = 0 & ω& = 0 using the Eule s equatins f tin, 1 tan θ = θ = 6.6 ) What t be deteined: Cuple ent,, M M, M M M = I ω& = I ω& ( I I ) ω ω = 0 ( I I ) ω (0) = 0 ( I I ) ω ω = 0 ( I I )(0) ω = 0 M = I ω& ( I I ) ω ω = 0 ( ) 10 3 (100sin 6.6)(100cs 6.6) = (0.877) 10 3 (100) = 8.77 N (Out f the plane) (nswe)

6 5. The clinde has a ass f 30 kg and is unted n an ale that is suppted b beaings at and. If the ale is tuning at ω = ( 40 ˆj ) ad/s, deteine the vetical cpnent f fce ( and ) acting at the beaings at this instant. Neglect the ale s ass. Sl) Step 1. Chse a ppe ving efeence sste ' ' ' - Oigin at - Mving with the ale ( Ω = ω ) - Pincipal aes Step. Find all fces and ents acting n the sste (Daw a fee-bd diaga), g Step 3. Use the equatins f tin (Fces and Ments) (1) F F = a cp: + = 0 (1) cp: = 0 = 0 (N inteest) + + = cp: 30 (9.8) = 0 () () Using the Eule s equatins f tin f the pincipal aes f efeence, ' ' ' ' cp: ' cp: ' cp: M = & ω ω ω ' I ' ' ( I ' I ' ) ' ' M = & ω ω ω ' I ' ' ( I ' I ' ) ' ' M & ω ω ω ' = I ' ' ( I ' I ' ) ' ' With espect t the pincipal aes f efeence, ' ' ' 1 I ' = I ' = (30)[3(0.5) + (1.5) ] = 6.09 kg 1 1 and I ' = (30)(0.5) = kg (F a clinde)

7 Then, ngula velcit ( ω, ω, ω ) and angula acceleatin ( ω&, 1. Deteine ω abut ' ' ' ω // (alng the ale) & 0.5 tan θ = 1.5 θ = ω&, ω& ) ω ' = 0 (N cpnent), ω ' = ω csθ = 40 cs18.43 = ad/s, ω = 40sin18.43 = 1.65 ad/s ', ω θ. Deteine & ω abut ' ' ' Since Ω = ω, then & ω (ω) = & (Independent t the cdinate sste) Thus, & ω in XYZ = 0 = (ω & ω& = 0, ω& = 0, ω& = 0 ) Then, the (Eule s) equatins f tin bece; ( )(1) + ( )(1) = 0 ( )( 37.95)( 1.65) = 474 (3) X X ( 1)sin (1)sin18.43 = (because ω 0 ) (4) X X ( 1) cs18.43 (1) cs18.43 = (because ω 0 ) (5) ' = ' = inseting Eq. (1) ( = ) int Eq. (4) = 0 = slving Eq. () + Eq. (3), (Eq. (): = & Eq. (3): + = 474 ) + = 1384 N and = 1090 N (NSWER)

8 6. The cnical pendulu cnsists f a ba f ass and length L that is suppted b the pin at its end. If the pin is subjected t a tatin ω, deteine the angle θ that the ba akes with the vetical as it tates. ls deteine the cpnents f eactin at the pin. Sl) Let s suppse : Pincipal aes Ments f inetia I 1 = I L & I = 0 3 = ω Z θ Y ngula velcit ω = ω cs θ ˆj + ω sinθ kˆ g X F the tatinal equatins f tin (Ment equatins), L 1 M = g( sinθ ) = I & ω ( I I ) ωω = 0 (0 L )( ω csθ )( ω sinθ ) 3 M = M = 0 (because & ω = 0 and ω = 0 ) Then, g 1 = Lω csθ 1 3g θ = cs ( ) 3 Lω F the tanslatinal equatins f tin (Fce equatins), (Use a fied XYZ fae) X cpnent: F = = ( a ) = 0 (N ccel. f alng X) Y cpnent: X X X FY = Y = ( a ) Y = ω ( ω : (whee L = sinθ ) Centipetal ccel. f ) Z cpnent: F = g = ( a ) = 0 (N ccel. f alng Z) = 0, X Z Z 1 Y = Lsinθω, Z = g Z L/ θ ω

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