Physics 106 Common Exm 1, Spring, 2004 Signture Nme (Print): A 4 Digit ID: Section: Instructions: Questions 1 through 10 re multiple-choice questions worth 5 points ech. Answer ech of them on the Scntron sheet using #2 pencil. Answer ll of them s there is no penlty for guessing. You will need to do clcultions on the exm ppers for most of the questions nd you my use bcks of the exm ppers for extr spce. Questions A nd B re workout problems worth 25 points ech. Answer them on the exm booklet nd show ALL work, otherwise there is no wy to give prtil credit. Be sure your nme nd section number re on both the Scntron form nd the exm booklet. Also write your nme, id, nd section t the top of ech pge with long nswer questions A nd B on them. You my bring nd use your own formul sheet, using one side of n 8.5 x 11 sheet or two sides of 5x8 crd. A defult formul sheet is lso provided (see finl pge of this booklet). Mke sure to bring your own clcultor: shring of clcultors is not permitted nd will need it. As you know, NJIT hs zero-tolernce policy for ethics code violtions. Students re not to communicte with ech other once the exm hs strted. All cell phones, pgers, or similr electronic devices should be turned off. If you hve questions or need something cll your proctor or instructor. 1. The drwing shows the blde of chin sw. The rotting sprocket tip t the end of the guide br hs rdius of 25 cm. The ngulr speed of the sprocket tip is 12 rd/s. Find the liner speed of chin link t point A. A) 4 m/s B) 5 m/s C) 1 m/s D) 2 m/s E) 3 m/s 2. A disk with rdius R= 3.0 m. is spinning bout its center. Initilly the disc hs n ngulr velocity of 150 rev/min, nd is slowing down uniformly t rte of 6.0 rd/s 2. By the time it stops spinning, the totl number of revolutions the disk will mke is: A) 7.1 B) 9.3 C) 1.1 D) 3.3 E) 5.5 3. Two identicl wheels re spinning, but wheel B is spinning with twice the ngulr velocity of wheel A nd its rdius is lso twice s lrge s the rdius of wheel A. The rtio of the rdil ccelertion of point on the rim of B to the rdil ccelertion of point on the rim of A is: A) 1/8 B) 1/4 C) 1 D) 2 E) 8 Pge of 6 1
4. Five equl 2.0-kg point msses re rrnged in the x-y plne s shown. They re connected by mssless sticks to form rigid body. The distnce is 4 m. Find the rottionl inerti in kg.m 2 bout n xis pqrllel to the z - xis through point P. The result is: y A) 128 B) 512 C) 288 D) 256 E) 192 x P 5. A uniform rod of mss M = 4 kg nd length L = 3.0m. is pivoted bout n xis perpendiculr to the rod nd 50 cm. from it's left end. Find the rottionl inerti bout this xis (in kg.m 2 ) A) 7.0 B) 3.0 C) 36.0 D) 5.0 E) 11.0 6. A force F is pplied to the edge of the wheel s shown t point P. The torque tht it produces bout n xis of rottion through the center of the wheel is: F f P R A) FR B) FRsin(f) C) Fsin(φ) D) Rsin(φ) E) FRcos(φ) 7. The dumbell in the figure consists of uniform rod fstened to two. msses ttched to ech end. The rottionl inerti of the rod bout n xis perpendiculr to the plne of the figure through point "O" is I ROD = 8.0 kg.m 2. Ech mss is 2.0 kg locted 2.0 meters from the center point "O" of the rod. Wht force F must be pplied to one end (perpendiculr to the rod) to give the system n instntneous ngulr ccelertion of 4.0 rd/s 2 bout the center? A. 32 N B. 48 N C. 64 N D. 27 N E. 18 N 2 kg O 2 kg F Pge of 6 2
8. A disk drive with rottionl inerti 4.0x10-3 kg.m 2 is to ccelerte uniformly from rest to n ngulr velocity of 7200 rev/min in 10 sec. The torque tht its motor must provide to cuse this ccelertion is: A) 0.30 N.m B) 0.72 N.m C) 3.30 N.m D) 180 N.m E) 2.9 N.m 9. Wht vlue of the ngle f would cuse the ngulr ccelertion to be zero in the digrm below. The forces F 1 nd F 2 ct on thin rigid rod pivoted t the rottion xis shown. The rottion xis is perpendiculr to the pge. A. 1.0 rd B. 0.0 rd C. 90 0 D. 45 0 E. none of these F 1 = 30 N. 50 cm 100 cm xis F 2 = 15 N. f 10. A 12 kg block hngs on cord tht is wrpped round the rim of flywheel of rdius 0.25 m. The rottionl inerti of the flywheel bout horizontl xis is 0.60 kg.m 2. When the block is relesed, the cord unwinds with no slipping, the system ccelertes, nd the tension in the cord is no longer equl to the weight of the block hnging from it. Find the ccelertion of the block: A) 49 m/s 2 B) 9.8 m/s 2 C) 5.4 m/s 2 D) 4.9 m/s 2 E) 0.02 m/s 2 12kg Pge of 6 3
Nme (Print) 4 Digit ID Show ll work for the following problem Section A A. (25 points) A smll bll of mss 2.5 kg is ttched to one end of 4.0-m-long mssless rod nd the other end of the rod is ttched to pivot s shown. The br is relesed from rest when it is horizontl t t = 0, fter which it swings down due to grvity. The sketch shows the br t time when it mkes n ngle θ with the verticl nd lso when t = 0. ) Clculte the net torque nd ngulr ccelertion of the br t the instnt fter it is relesed. b) Find the net torque on the br nd its ngulr ccelertion when θ = π/6. Show the moment rm you re using on the sketch. c) Find the ngulr velocity ω when θ = 0; tht is, when the br hs swung so tht it is momentrily verticl. There is no friction in the pivot. Note tht the ngulr ccelertion is not constnt in this problem. d) Find the ngulr ccelertion when θ = 0. Describe wht hppens to the torque nd ngulr ccelertion s the br swings pst this point. L Answers: ) t = - 98 nm, = -2.5 rd/s 2 b) t = - 49 nm, = -1.2 rd/s 2 c) w = - 2.2 rd/s d) = 0 since the moment rm is proportionl to sin(q), which vnishes for q = 0. Afterwrds, both nd t reverse direction nd become counter-clockwise. The ngulr velocity remins negtive (CW) until the br hs swung to its most leftmost position. q mg Pge of 6 4
Nme (Print) 4 Digit ID Show ll work for the following problem: Section A B. (25 points) Two msses re ttched by cord to pulley s shown. The cord does not stretch nd it cn not slip where it touches the pulley, but there is no other frictionl force. The rottionl inerti of the pulley is 5.0 kg.m 2. The rdius R =.75 m s shown. ) Drw free-body digrms nd use Newton's second lw to get the equtions tht describe the motion of mss A nd mss B. b) Drw the free-body digrm nd get the rottionl second lw eqution for the pulley. Include the forces nd torques cting on it. c) You need 2 more equtions to solve for the 5 unknowns in your equtions bove. Wht re they nd why re they justified? d) Clculte the ngulr ccelertion of the pulley. Ans - 1.1 rd/s 2 CW 0.75 m e) Clculte the ccelertion of mss B. Ans: - 0.86 m/s 2 (down) A T A M A = 6 kg T B A B M B = 8 kg Pge of 6 5
Physics 106: PHYSICS I FORMULAS - Common Exm 1 360 degrees = 2π rdins = 1 revolution. s = rθ v t = rω t = rα c = r =v t 2 /r = ω 2 r tot 2 = r 2 + t 2 ω = ω o + αt θ f - θ o = ω o t +½αt 2 ω f 2 - ω o 2 = 2α(θ- θ o ) θ - θ o = ½(ω+ ω o )t KE rot = 1/2Iω 2 I = Sm i r i 2 I point = mr 2 I hoop = MR 2 I disk = 1/2 MR 2 I sphere = 2/5 MR 2 I shell = 2/3 MR 2 I rod (center) = 1/12 ML 2 I rod (end) = 1/3 ML 2 I p = I cm + Mh 2 τ = force x moment rm = Frsin(φ) = r x F t net = Σt= Ι F net = ΣF = m W tot = K = K f - K I W = τ net θ P = τ.ω (τ constnt) K = K rot + K cm E mech = K + U E mech = 0 (isolted system) v cm = ωr (rolling, no slipping) l = r x p p = mv L = S l i t net = dl/dt L = Iω l point mss = mrvsin(φ) For isolted systems: t net = 0 L is constnt L = 0 L 0 = ΣI 0 ω 0 = L f = ΣI f ω f Equilibrium: Σforces = 0 nd Σtorques = 0, If net force on system is zero, then the net torque is the sme for ny chosen rottion xis. CG definition: point bout which torques due to grvity lone dd to zero. Physics 105: F g = mg g = 9.8 m/s 2 1 m = 100 cm = 1000 mm, 1 kg = 1000 g v = v o + t x - x o = v o t + ½t 2 v 2 - v 2 o = 2(x - x o ) x - x o = ½(v + v o )t F net = m ΣF = m F s,mx = µ s N F k = µ k N incline: W mgx = mgsinθ W mgy = mgcosθ F r = m r = mv 2 /r r = v 2 /r f = 1/T T = (2πr/v) Impulse: F vr t = mv f - mv I Momentum conserved if net impulse = 0. Then (Smv) initil = (Smv) finl Work: W = Fd(cosθ), W grv = - mg(y-y 0 ), W spring = -1/2k(x 2 -x 2 0 ), W frict = -F k d, W tot = K f - K i U g = mg(y-y 0 ), spring: F = -kx, U s = 1/2kx 2, KE = 1/2mv 2 W nc = K f - K I + U gf - U gi + U sf - U si or K I + U gi +U si + W nc =K f + U gf + U sf Mss center: X cm = S(x i m i )/Σm i, similrly for Y cm, Z cm Vectors: Components: x =.cos(θ) y =.sin(θ) = x i + y j = sqrt[ x 2 + y 2 ] θ = tn -1 ( y / x ) Addition: + b = c implies c x = x + b x, c y = y + b y Dot product: o b =.b.cos(φ) = x b x + y b y + z b z unit vectors: i o i = j o j = k o k = 1, i o j = i o k = j o k = 0 Cross product: x b =.b.sin(φ) x b = - b x, x = 0 lwys. c = x b is perpendiculr to -b plne i x i = j x j = k x k = 0, i x j = k j x k = i k x i = j etc. Pge of 6 6