Unformatted text preview: Answer, Key Homework 8 Rubin H Landau 1 This print-out should have 14 questions. the center is: Z Check that it is complete before leaving the Icm = r2 dr printer. Also, multiple-choice questions may continue on the next column or page: nd all ZR m choices before making your selection. = r2 r dr 2 , R2 l Rout in R We are back in order let's hope. Some Z 2 Zl solutions may be found on the class home d dz page. 0 0 4 4 Rout , Rin 2l m Rolling of a Cylinder = 2 2 4 Rout , Rin l 11:01, trigonometry, numeric, 1 min. m R2 + R2 004 = 2 out in An m = 3:34 kg hollow cylinder with Rin = 0:3 m and Rout = 0:5 m is pulled by a horizon- By the parallel axis theorem, the moment of tal string with a force F = 22:8 N, as shown inertia about the ground is: in the diagram. 2 I = Icm + mRout 2 2 = m 3Rout + Rin 2
out in m F Rin Rout What must the magnitude of the force of friction be if the cylinder is to roll without slipping? Correct answer: 4:34286 N. Explanation: Suppose the cylinder has a length of l. Then m density of the cylinder is = the 2 2 Rout , Rin l . The moment of inertia about From the force equation, we have: F + f = ma: From the torque equation, we have: F 2Rout = I = I Ra : out Solving this pair of equations, we get: 2 2mRout , 1 f =F I 23:34 kg0:5 m2 , 1 = 22:8 N 1:4028 kgm2 = 4:34286 N In what direction is the frictional force? 005 1. To the left 2. To the right correct 3. Force is zero Explanation: 006 The frictional force calculated has the same sign as the applied force, so it must be in the same direction as F , i.e. to the right. Answer, Key Homework 8 Rubin H Landau 2 From the perspective of the ground and What is the acceleration of the cylinder's cennoting that the acceleration is constant, the ter of mass? distance traveled by the disk before the pure Correct answer: 8:1266 m=s2. rolling occurs is, Explanation: From the pair of equations above, we can solve for a: 2 2FRout s = 1 a t2 a= I 2 1 g R !0 2 222:8 N0:5 m2 : =2 = 1:4028 kgm2 3 g R2 ! 2 = 8:1266 m=s2 = 18 0 g Algorithm 0:1 m2 =s2 = 0:187064319:8rad=s2 r=0:3 m 0:3 1 18 0: 9 m 0:4 R=0:5 m 0:6 2 = 1:11296 m : 2 m=3:34 kg 5 3 20 F =22:8 N 50 4 , I =0:5 m 3:0 R2:0 +r2:0 5 Algorithm i = 0:01 m=cm m 1 hcm , 2:0 2: 0 =0:5 h3:34i 3:0 h0:5i +h0:3i 2 g = 9:8 m=s 2 2 =1:4028 kgm r = 18:7 cm 10 3 , 30 2 2:0 2:0 10 hkgm i=hi hkgi hi hmi +hmi units !0 = 19 rad=s 20 4 0:05 2:0 F R2:0 a= 6 = 0:0643 0:2 5 I mi R = r hcm 6 2:0 = 2:0 h22::8i h0:i5i = h18:7i h0:01i h1 4028 = 0:187 m =8:1266 m=s2 hmi = hcmi hm=cmi units hi hNi hmi2:0 2 hm=s i= hkgm2i units 7 t = 3R !0g :0 2:0 0: i f =F 2:0 mIR ,1 7 = 3:0hh0187i hi199:8i :0643 h 2:0 h3:34i h0:5i2:0 ,1 = 1:87948 s =h22:8i h1:4028i i si hsi = hmhihhrad=2i units =4:34286 N hi m=s i2:0 2:0 2:0 hNi=hNi hi hkgi hmi ,hi units 2 s = R :0!0g 8 hkgm 18 2:0 h 2:0 h = 180::0187:i064319i :8i Contacting a Surface h0 i h9 11:01, advanced, numeric, 1 min. = 1:11296 m 008 i2:0 si2:0 hmi = hmhi hihhrad=2i units Determine the distance the disk travels before m=s pure rolling occurs. Correct answer: 1:11296 m. Explanation: 009 l1 m l1 M2 Answer, Key Homework 8 Rubin H Landau 2 l2 2. 12 m s 2 3. 13:5 kg s2m M1 4. 18 Nkgm l2 5. 9 Nkgm
m 3 Explanation: A rod of negligible mass is pivoted at a point that is o -center, so that length `1 is diferent from length `2 . the gures above show two cases in which masses are suspended from the ends of the rod. In each case the unknown mass m is balanced by a known mass, M1 or M2, so that the rod remains horizontal. What is the value of m in terms of the known masses? The angular momentum is L = mvr = 2 kg3 m=s4 m 2 = 24 kg s m :
11:01, calculus, numeric, 1 min. 1. M1 + M2 5. M12M2 Explanation: 2. M1 + M2 2 3. M1M2 p 4. M1M2 correct A 7:9 kg mass is attached to a light cord, which is wound around a pulley. The pulley is a uniform solid cylinder of radius 11:6 cm and mass 1:89 kg. What is the resultant net torque on the system about the center of the wheel? Correct answer: 8:98072 kg m2=s2 . Mass on Solid Cylinder 011 Explanation: The balance in the rst case requires m`1 = M1`2. And the balance in the second case requires M2`1 = m`2. Cancel `1 and `2 from the above equations. So m2 = M1M2 , i.e. p m = M1 M2. A 2 kg object moves in a circle of radius 4 m at a constant speed of 3 m s. A net force of 4.5 N acts on the object. What is the angular momentum of the object with respect to an axis perpendicular to the circle and through its center? 2 1. 24 kg s m correct The net torque on the system is the torque by the external force, which is the weight of the mass. So it is given by: = r F sin = r m g sin 90 = r m g = 0:116 m 7:9 kg 9:8 m=s2 = 8:98072 kg m2=s2: When the falling mass has a speed of 5:47 m=s, the pulley has an angular velocity of v=r. Determine the total angular momentum of the system about the center of the wheel. Correct answer: 5:61233 kg m2=s. 010 012 Explanation: The total angular momentum has two parts, one of the pulley and one of the mass. Answer, Key Homework 8 Rubin H Landau 2:0 So it is I = 1:0Mru 2:0 ~r jLj = j~ m ~ + I ~ j v! ih0 2:0 v = 1:0h1:892:0 :116i = r m v + 1 M r2 r = r m + M v 2 2 = 0:0127159 kg m2 = 11:6 cm 7:9 kg + 1:89 kg v = 1:02602 kg m v hkg m2i = hihkgihmi2:0 2 hi v = 1:02602 kg m 5:47 m=s = 5:61233 kg m2=s : !=r u 47 Mass on Solid Cylinder = hh05::116ii 11:02, calculus, multiple choice, 1 min. Using the fact that = dL=dt and your result from the previous part, calculate the acceleration of the falling mass. Correct answer: 8:75297 m=s2. 4 9 units 10 013 Explanation: Use the torque-angular momentum relation, we have d = dL = dt 1:02602 kg m v = 1:02602 kg m a ; dt Solving for acceleration: a = 1:02602 kg m 8:98072 kg m2=s2 = 1:02602 kg m = 8:75297 m=s2: Algorithm m = hcmi = 0:01 mcm m = 7:9 kg 4 8 8 r = 11:6 cm 15 m ru = rhcmi = h11:6ih0:01i = 0:116 m hmi = hcmihm=cmi M = 1:89 kg 1 3 2 g = 9:8 m=s = rumg = h0:116ih7:9ih9:8i = 8:98072 kg m2=s2 2 hkg m2=s2 i = hmihkgihm=s i 5 v = 5:47 m=s 10 1 2 3 4 units 5 6 7 units 8 = 47:1552 s,1 s hs,1i = hm=i i units hm L1 = rumv 11 = h0:116ih7:9ih5:47i = 5:01271 kg m2=s hkg m2=si = hmihkgihm=si units L2 = I! 12 = h0:0127159ih47:1552i = 0:599621 kg m2=s hkg m2=si = hkg m2ihs,1i units L = L1 + L2 13 = h5:01271i + h0:599621i = 5:61233 kg m2=s hkg m2=si = hkg m2=si + hkg m2=si units 14 b= L v 61233 = h5h:5:47i i = 1:02602 kg m m2 units hkg mi = hkgm=s=si hi v a= L 15 : = h8:98072ih5i47i h5:61233 = 8:75297 m=s2 22 hm=s2i = hkg hm =s 2ihsm=si kg m = i units 11:03, trigonometry, multiple choice, 1 min. Child on a MerryGoRound Answer, Key Homework 8 Rubin H Landau 5 014 = h200i + h21i h2:5i2:0 A playground merry-go-round of radius 2:5 m = 331:25 kg m2 2 has a moment of inertia 200 kg m and is hkg m2i = hkg m2i + hkgi hmi2:0 units rotating at 9:5 rev=min. A child with mass 21 kg jumps on the edge of the merry-go6 !2 = !1II1 2 round. 5i h What is the new moment of inertia of the = h9h:331:200ii 25 merry-go-round and child, together? Correct answer: 331:25 kg m2. = 5:73585 rev=min Explanation: h2 hrev=mini = hrev=minimkg m i units The moment of inertia will be the combihkg 2i nation of the individual moments of inertia of each component. Imerry,go,round + Ichild = Itotal Child on a MerryGoRound 11:03, calculus, numeric, 1 min. 015 Explanation: Basic Concepts:
X Assuming that the boy's initial speed is negligible, what is the new angular speed of the merry-go-round? Correct answer: 5:73585 rev=min. A bicycle wheel of mass m rotating at an angular velocity ! has its shaft supported on one side, as shown in the gure. When viewing from the left, one sees that the wheel is rotating in a counterclockwise manner. The distance from the center of the wheel to the pivot point is b. We assume the wheel is a hoop of radius R, and the shaft is horizontal.
b 018 ~ L = const
ω τ The net angular momentum of the system remains constant, therefore, from conservation of the angular momentum we have: I1 !1 = I1 + m R2 !2 And 1 !2 = !1 I +Im R2 1 9:5 rev=min 200 kg m2 = 200 kg m2 + 21 kg 2:5 m2 = 5:73585 rev=min W=mg The magnitude of the angular momentum of the wheel is given by 1. 1 m R2 ! 4 Algorithm 2. m R2 !2 3. 1 m R2 !2 2 4. 1 m R2 !2 4 R = 2:5 m 1::5 25 I1 = 200 kg m2 100 300 5 !1 = 9:5 rev=min 15 m = 21 kg 20 35 I2 = I1 + m R2:0 1 2 3 4 5 Answer, Key Homework 8 Rubin H Landau 6 5. 1 m R2 ! 2 6. m R2 ! correct Explanation: Solution: Basic Concepts: ~ ~ = dL
Top view 11:05, calculus, numeric, 1 min. The direction of precession as viewed from the top is: Precession 020 1. along the direction of rotation of the wheel 2. counterclockwise correct 3. clockwise 4. opposite to the direction of rotation of the wheel Explanation: Algorithm hdegi = 57:2958 deg=rad rad
∆L L+ ∆L ∆φ τ The magnitude of the angular momentum of the wheel, L, is L = I ! = m R2 !; since the moment of inertia of the wheel, I , is m R2 . Given: the mass 3 kg, the angular velocity 15 rad=s, the axil length b = 0:5 m, and the radius of the wheel R = 0:49 m. Find the precession angle in the time interval t = 1:1 s. Correct answer: 85:7488 . From the gure below, we get = L . L Using the relation, L = t, where is the magnitude of the torque, mg b, we get = L L = Lt b = m gR2t m! b = gR2t ! 9:8 m=s20:5 m1:1 s = 0:49 m215 rad=s = 1:4966 rad = 85:7488 : From the gure, we can see the direction of precession is counterclockwise. 019 Explanation: g = 9:8 m=s m = 3:0 kg ! = 15 rad=s 10 15 0:4 b = 0:5 m 0:6 R = 0:49 m 0::4 1 0 6 t = 1:1 s 2 = Rgbt! 2:0 : ih 1: = h908490:250ih151i h : i:h i = 1:4966 rad hm= 2 hradi = hmi2s:0ihmihsi hrad=si degi deg = hrad = h1:4966ih57:2958i = 85:7488 h i = hradihdeg=radi 1 2 3 4 5 6 7 8 units 9 units ...
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