= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5.

Transcription

1 Seat: PHYS 500 (Fall 0) Exa #, V 5 pt. Fro book Mult Choice 8.6 A tudent lie on a very light, rigid board with a cale under each end. Her feet are directly over one cale and her body i poitioned a hown. The left cale read 60 lb and the right cale read 00 lb. What i the tudent weight? (a) 60 lb 5 pt Nae: (b) 80 lb (c) 00 lb (d) 60 lb. Fro book Concep Que 9.3 A 0. kg platic cart and a 0 kg lead cart can roll without friction on a horizontal urface. Equal force are ued to puh both cart forward for a tie of, tarting fro ret. After the force i reoved at t =, i the oentu of the platic cart greater than, le than, or equal to the oentu of the lead cart? Explain. Becaue they have the ae force acting on the for the ae aount of tie, they ut have the ae ipule. The ipule alo equal the change in oentu. They both tart with 0 oentu. Therefore, their final oenta are the ae. 5 pt 3. Fro book HWK Prob 3.37 You are on the edge of a erry-go-round (take your a to be 70 kg and that of the erry-go-round to be 500 kg) that pin at 8 rp. The diaeter of the erry-go-round i 4.6. a) What are the erry-go-round period (in ) and frequency in (rev/)? b) What i your peed? c) What i your acceleration? (a) The converion fro rp (rotation per inute) to frequency in (rev/) i divide by 60. The period i T = /f. (b) The object peed i the circuference divided by the period: v = πr/t = πdiaeter/t. (c) The acceleration i the centripetal acceleration a = v /r. 8 rev rev f= = 0.30 = π 4.6 (4.34 /) v= = 4.34 a= = 8.7 (V ) rev rev = 0.7 = π 4.4 (3.69 /) v= = 3.69 a= = f= rev rev = 0.37 = π 4.8 (5.53 /) v= = 5.53 a= = (V ) f=

2 0 pt 4. Fro book HWK Prob 6.49 A 55.0 g ball i on a.0 long tring o that the ball ove at contant peed in a horizontal circle of radiu 49.0 c. a) What i the tenion in the tring? b) What i the ball angular velocity? To ee how to do thi proble, you hould ake a free body diagra like the one above. You then need to ue Newton nd law. The x-coponent of the net force i T co θ + 0 = ax and the y-coponent of the net force i T in θ g = ay. The x-coponent of the acceleration ut be v /r for the ball to go in a circle. The y-coponent of the acceleration i 0 ince the ball ove in a horizontal circle. For part b), you can find the angular velocity fro ω = v/r. Latly, you can find the angle fro the triangle ade fro the tring and the radiu of the circle: θ = co (r/ ). θ = co (0.49) = 60.7 v = 0.68 N co(60.7 ) 0.49 = kg θ = co (0.54) = 57.3 v = N co(57.3 ) 0.54 = kg θ = co (0.59) = 53.8 v = N co(53.8 ) 0.59 = kg kg 9.8 / = 0.68 N in(60.7 ) v =.64 ω=.64 / = 3.35 rad/ 0.49 (V ) kg 9.8 / = N in(57.3 ) v =.84 ω=.84 / = 3.4 rad/ 0.54 (V ) kg 9.8 / = N in(53.8 ) v =.06 ω=.06 / = 3.48 rad/ 0.59

3 5 pt 5. Fro book Mult Choice 6.9 Two planet orbit a tar. Planet ha orbital radiu r and planet ha r = 4r. Planet orbit with period T. Planet orbit with period (a) T = T (b) T = T (c) T = 4T (d) T = 8T 5 pt 6. Fro book Concep Que 8. An object i acted upon by two (and only two) force that are of equal agnitude and oppoitely directed. I the object necearily in tatic equilibriu? Explain and give an exaple. No, it in t necearily in tatic equilibriu. The reaon i that the object can have a ize. Generalized exaple: if the two force don t act on a line, then the object will have an angular acceleration. Specific exaple: puh up on the edge of your book with a force equal to the weight (the other force i the weight fro gravity), the book will pin. 5 pt 7. Baed on book Prob 7.0 A thin rod ha a length of.0, a pivot-point 50.0 c fro the left end, and alot no a. The rod i at an angle 30.0 above the horizontal. A force of 3.80 N act on the right end in the +y-direction. A force of 5.3 N act on the left end in the x-direction. What i the torque on the rod? You need to add the torque fro the two force uing τ = rf in(ϕ) for each force. Define θ to be the angle given in the proble and the two torque a in the figure below. ϕ = θ and ϕ = 90 θ. Note τ i negative and τ i poitive. τ = τ + τ = N in(30 ) N in(60 ) =.77 N (V ) τ = τ + τ = N in(40 ) N in(50 ) =.9 N (V ) τ = τ + τ = N in(30 ) N in(60 ) = 0.55 N

4 0 pt 8. Fro book HWK Prob 9.6 In a daring recue in pace, Jeica (a of 77.0 kg with uit) i traveling at.60 / in a direction 33.0 above the +x-axi. Meanwhile, Michael (a 98.0 kg with uit) i traveling in the y-direction with a peed of.80 /. They collide and hold onto each other. Give the velocity of the reulting Jeica-Michael unit. Thi i a conervation of oentu proble. You need to find the oentu before the colliion and et it equal to the oentu after the colliion. You ut ue coponent to get the anwer. Thi give two equation: M j v x,j + M v x, = (M j + M )v x,j M j v y,j + M v y, = (M j + M )v y,j All of Michael velocity i in the y-direction. The coponent of Jeica velocity are v x,j = v j co(θ) and v y,j = v j in(θ). v x,j =.6 co(33 ) =.8 v y,j =.6 in(33 ) =.4 v y,j = v x,j =.6 co(33 ) =.8 v y,j =.6 in(33 ) =.4 v y,j = v x,j =.6 co(36 ) =.0 v y,j =.6 in(36 ) =.53 v y,j = v x,j = 77 kg.8 / 75 kg = 0.96 / 77 kg.4 / + 98 kg(.8 /) 75 kg v x,j = 83 kg.8 / 8 kg = 0.99 / 83 kg.4 / + 99 kg(.7 /) 8 kg v x,j = 77 kg.0 / 76 kg = 0.9 / 77 kg.53 / + 99 kg(.9 /) 76 kg = 0.38 / (V ) = 0.8 / (V ) = 0.40 /

5 5 pt 9. Fro book Prob 7.6 Three coin are centered on three of the corner of a quare a hown. The length of a ide i L = c. The horizontal poition of the center of a i (a) 0 c (d) 6 c 5 pt (b) 3 c (e) 9 c (c) 4 c (f) c 0. Fro book Concep Que 3.4 You are driving your car in a circular path on level ground at a contant peed of 0 ph. At the intant you are driving north, and turning left, are you accelerating? If o, toward what point of the copa (N, W, S, E, NW, SW,...) doe your acceleration vector point? If not, why not? Explain your anwer. The car i accelerating toward the center of the circle. Since the car i oving north but turning left, the center i to the wet which i the direction of the acceleration. 5 pt. Fro book HWK Prob 9.35 In cla, Prof Robicheaux at on a chair that wa free to pin. He tarted with the weight held in outtretched ar and finihed with the weight pulled into hi body. With weight in outtretched ar the oent of inertia wa 9.00 kg while it wa 4.00 kg with the weight pulled in. If he had an angular peed of 6.60 rad/ with hi ar outtretched, what wa the angular peed with hi ar pulled in? Thi i a conervation of angular oentu proble becaue there i no external torque on thi yte. Thi ean Lf = Li. You can figure out the final angular velocity uing If ωf = Ii ωi. ωf = Ii ωi 9 kg 6.6 rad/ = = 4.9 rad/ If 4 kg (V ) ωf = Ii ωi kg 4.4 rad/ = = 9.68 rad/ If 5 kg (V ) Ii ωi 8 kg 4.9 rad/ ωf = = = 3. rad/ If 3 kg

6 0 pt. Fro book Prob 7.64 The 3.0 kg, 40.0-c-diaeter dik i pinning at rp. How uch friction force ut the brake apply to the ri to bring the dik to a halt in 3.0? The brake give a torque to the dik oppoite the angular velocity. Uing the oent of ineria of the dik you can find the angular acceleration. Strategy: fro the angular velocity and the tie to halt the dik, find the angular acceleration you need: ωf = ωi +α t. Uing the angular acceleration and the oent of inertia you can find the torque needed: τ = Iα. Fro the torque and the radiu of the dik you can find the force the applied by the brake: τ = rf in ϕ with ϕ = 90 and the becaue the torque i in the clockwie direction. rot π rad in = 3.4 rad/ in rot 60 0 = 3.4 rad/ + α3 α = 0.5 rad/ ω = 300 I = MR = 3 kg (0.0 ) = 0.06 kg τ = 0.06 kg ( 0.5 rad/ ) = 0.63 N 0.63 N F= = 3.5 N (V ) 0. rot π rad in = 3.4 rad/ in rot 60 0 = 3.4 rad/ + α4 α = 7.85 rad/ ω = 300 I = MR = 4 kg (0.5 ) = 0.5 kg τ = 0.5 kg ( 7.85 rad/ ) = 0.98 N 0.98 N F= = 3.93 N (V ) 0.5 rot π rad in = 3.4 rad/ in rot 60 0 = 3.4 rad/ + α5 α = 6.8 rad/ ω = 300 I = MR = 5 kg (0.30 ) = 0.5 kg τ = 0.5 kg ( 6.8 rad/ ) =.4 N.4 N F= = 4.7 N 0.3

7 Equation Baic Matheatic Forula in θ = O/H co θ = A/H tan θ = O/A H = O + A A circ = πr C circ = πr V ph = 4π 3 r3 A ur of ph = 4πr x = b ± b 4ac a Chapter x = x f x i v x = x/ t ā x = [(v x ) f (v x ) i ]/(t f t i ) = v x / t (v x ) f = (v x ) i +a x t x = [(v x) i +(v x ) f ] t x f = x i +(v x ) i t+ a x t (v x ) f = (v x ) i +a x x Chapter 3 A x = A co θ A y = A in θ A = A x + A y tan θ = A y /A x d = d f d i v av = d/ t a av = ( v)/ t (v y ) f = (v y ) i +a y t y = [(v y) i +(v y ) f ] t y f = y i +(v y ) i t+ a y t (v y ) f = (v y ) i +a y y f = /T v = πr/t a = v /R Chapter 4 F net = F + F + F = a F = F Chapter 5 w = g f ax = µ n f k = µ k n f r = µ r n θ = r ω av = θ f θ i t f t i = θ t Chapter 6 ω = (π rad)f v = ωr a = v r = ω r F g = G r N G = kg T = 4π GM r3 α av = ω f ω i t f t i = ω t Chapter 7 a t = αr x cg = x + x + x ω = ω i + α t θ = (ω i + ω) t θ = ω i t + α t ω = ω i + α θ τ = rf = r F = rf in ϕ I r τ = Iα vobj = ωr a obj = αr Fx = 0 Chapter 8 Fy = 0 τ = 0 Fp = k x

8 Chapter 9 p = v J = Favg t Favg t = p = v f v i v i + v i = v f + v f L Iω Chapter 0 τ = L t W = E = K + U g + U + E th + E che +... W = F d = (F co θ)d K tran = v K rot = Iω U g = gy U = kx P = E t P = F v e = what you get what you pay Chapter E th = W + Q e = Q H Q C Q H e ax = T C T H COP = Q c W in COP ax = T C T H T C COP = Q H W in COP ax = T H T H T C Chapter n = N/N A = M(in gra)/m ol p = F/A pv = Nk B T = nrt T C = T 73 T F = 9 5 T C + 3 L = αl i T V = βv i T Q = Mc T Qk = 0 Q = ±ML Q t = kat h T c L Q t = σae(t 4 T 4 0 ) ρ = V p = F A Chapter 3 p = p 0 +ρgd F B = ρ f V f g A v = A v p + ρv +ρgy = p + ρv +ρgy Chapter 4 F = kx U = kx T = π k f = T ω = πf x = A co(ωt) L v = Aω in(ωt) a = Aω co(ωt) T = π g Soe Propertie of Material Material denity Linear Expanion Coef Specific Heat Lead 000 kg/ C 8 J/(kg C) Iron/Steel 7900 kg/ C 448 J/(kg C) Copper 8900 kg/ C 387 J/(kg C) Aluinu 700 kg/ C 900 J/(kg C) Water 000 kg/ 3 N.A. 490 J/(kg C)