xu (rx575) – HW08 – macdonald – (55475) This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – ﬁnd all choices before answering. 001 (part 1 of 2) 10.0 points A record has an angular speed 19.9 rev/min. What is its angular speed? of Let : Lh Lm Mh Mm 1 = 2.79 m , = 4.54 m , = 58.4 kg , = 110 kg . and The hour hand makes 1 rev in 24 hours and the minute hand in 60 minutes. The moment of inertia of a thin rod about an axis through one end is M L2 I= , 3 so the total rotational kinetic energy is Correct answer: 2.08392 rad/s. Explanation: Let : ω = 19.9 rev/min . 2 Ih ωh2 Im ωm + 2 2 2 mh Lh 2 mm L2m 2 = ωh + ωm 6 6 (58.4 kg) (2.79 m)2 = 6 2 2 1h 2 π rad × 12 h 3600 s (110 kg) (4.54 m)2 + 6 2 2 2 π rad 1 min × 60 min 60 s Ktot = ω = (19.9 rev/min) 2 π rad rev 1 min 60 s = 2.08392 rad/s . 002 (part 2 of 2) 10.0 points Through what angle does it rotate in 1.55 s? Correct answer: 3.23008 rad. Explanation: Let : t = 1.55 s . θ = ωt = (2.08392 rad/s) (1.55 s) = 3.23008 rad . 003 10.0 points The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.54 m long and have masses of 58.4 kg and 110 kg, respectively. Calculate the total rotational kinetic energy of the two hands about the axis of rotation. Model the hands as long thin rods. Correct answer: 0.00115269 J. Explanation: = 0.00115269 J . Note: The rotational kinetic energy of the hour hand is negligible compared to the minute hand. keywords: 004 (part 1 of 2) 10.0 points A racing car travels on a circular track of radius 154 m, moving with a constant linear speed of 42.3 m/s. Find its angular speed. Correct answer: 0.274675 rad/s. Explanation: Let : v = 42.3 m/s R = 154 m . and xu (rx575) – HW08 – macdonald – (55475) The linear speed is v = Rω v 42.3 m/s ω= = = 0.274675 rad/s . R 154 m 2 A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 19840 rev/min. Find the drill’s angular acceleration. Correct answer: 485.43 rad/s2 . 005 (part 2 of 2) 10.0 points Find the magnitude of its acceleration. Correct answer: 11.6188 m/s2 . Explanation: With the car moving at a constant speed, there is no tangential acceleration, so the acceleration is purely radial: ar = v2 (42.3 m/s)2 = = 11.6188 m/s2 . R 154 m 006 10.0 points Initially a wheel rotating about a ﬁxed axis at a constant angular deceleration of 0.4 rad/s2 has an angular velocity of 0 rad/s and an angular position of 8.3 rad. What is the angular position of the wheel after 3.5 s? Correct answer: 5.85 rad. Explanation: Let : Explanation: Let : t = 4.28 s and ωf = 19840 rev/min . Since ω0 = 0, ωf − ω0 ωf α= = t t 19840 rev/min 2 π 1 min = · · 4.28 s rev 60 s = 485.43 rad/s2 . 008 (part 2 of 2) 10.0 points Throughout what angle does the drill rotate during this period? Correct answer: 4446.15 rad. Explanation: θ = ω0 t + 1 α t2 2 1 (485.43 rad/s2 ) (4.28 s)2 2 = 4446.15 rad . =0+ α = −0.4 rad/s2 , ω0 = 0 rad/s , θ0 = 8.3 rad , and t = 3.5 s . From kinematics, 1 2 α t + θ0 2 = (0 rad/s) (3.5 s) 1 + (−0.4 rad/s2 ) (3.5 s)2 + 8.3 rad 2 = 5.85 rad . 009 (part 1 of 4) 10.0 points Consider a rod of length L and mass m which is pivoted at one end. An object with mass m is attached to the free end of the rod. θ = ω0 t + 007 (part 1 of 2) 10.0 points m C L 20◦ m xu (rx575) – HW08 – macdonald – (55475) Determine the moment of inertia of the system with respect to the pivot point. The acceleration of gravity g = 9.8 m/s2 . Consider the mass at the end of the rod to be a point particle. 3 m L2 2 4 2. I = m L2 correct 3 1. I = 3. None of these. 5 m L2 3 13 m L2 5. I = 12 4. I = 6. I = L2 5 m L2 4 Explanation: The moment of inertia of the rod with respect to the pivot point is 7. I = Irod = 1 m L2 , 3 and the moment of inertia of the mass m with respect to the pivot point is Imass = m L2 . Thus the moment of inertia of the system is I = Irod + Im 1 4 = m L2 + m L2 = m L2 . 3 3 010 (part 2 of 4) 10.0 points The length C in the ﬁgure represents the location of the center-of-mass of the rod plus mass system. Determine the position of the center of mass from the pivot point. 7 1. C = L 8 3 5. None of these 1 L 2 Explanation: 6. C = 1 The center of mass of the rod is L, so the 2 center of mass of the rod-plus-mass system is 1 L+L m 2 C= m+m 1 1 3 = L+ L = L. 4 2 4 011 (part 3 of 4) 10.0 points The unit is released from rest in the horizontal position. What is the kinetic energy of the unit when the rod momentarily has a vertical orientation? 1 mgL 2 3 2. K = m g L correct 2 1. K = 3. K = 2 m g L 4. None of these 5. K = 5 mgL 2 6. K = m g L Explanation: The potential energy released can be computed in two ways: ∆U = ∆U CM of rod + ∆U mass L 3 = mg + mgL = mgL, 2 2 or 2. C = L 3 L correct 4 5 4. C = L 8 3. C = ∆U |rod+m = ∆U |CM of rod+m 3 3 L = mgL. = (2 m) g 4 2 xu (rx575) – HW08 – macdonald – (55475) 012 (part 4 of 4) 10.0 points Let the kinetic energy be KV at the vertical position and the moment of inertia of the system be I. Find the angular velocity of the system in terms of KV and I. 2 KV I r 2 KV 2. ω = correct I r KV 3. ω = I 1. ω = 4. None of these 4 ω2 r = α r √ ω= α and the angular velocity is ω = ω0 + α ∆t ω ω − ω0 = α= ∆t ∆t since ω0 = 0, so √ ω α 1 t= = =√ α α α 1 =p 4.9 rad/s2 = 0.451754 s . KV I r KV 6. ω = 2I Explanation: The rotational kinetic energy of the system is given by 1 KV = I ω 2 2 r 2 KV . ω= I 5. ω = 013 10.0 points A horizontal disk with a radius of 20 m rotates about a vertical axis through its center. The disk starts from rest and has a constant angular acceleration of 4.9 rad/s2 . At what time will the radial and tangential components of the linear acceleration of a point on the rim of the disk be equal in magnitude? Correct answer: 0.451754 s. Explanation: Let : r = 20 m and α = 4.9 rad/s2 . The tangential and radial accelerations are equal: ar = at 014 (part 1 of 2) 10.0 points Consider a 5 kg square which has its mass concentrated along its perimeter, with each side of length 6 m. d d What is the moment of inertia of the square about an axis perpendicular to the plane of the square at its center of mass? Use the parallel axis theorem and divide the square into parts. The moment of inertia of a rod 1 CM m d2 . rotated about its CM is Irod = 12 Correct answer: 60 kg · m2 . Explanation: Let : 5 kg and 4 ℓ = 6 m. mrod = By the parallel axis theorem, the moment of inertia of each rod about the center of the square is 2 d 1 1 rod 2 = mrod d2 . Isquare = mrod d +mrod 12 2 3 xu (rx575) – HW08 – macdonald – (55475) Since there are four sides to the square, the moment of inertia of the square about its center of mass is 1 CM rod 2 Isquare = 4 Isquare = 4 mrod d 3 4 5 kg 4 ) (6 m)2 = mrod ℓ2 = ( 3 3 4 = 60 kg · m2 . 015 (part 2 of 2) 10.0 points d d What is the moment of inertia of the square about an axis perpendicular to the plane of the square at one of its corners? Correct answer: 150 kg · m2 . Explanation: The distance from the center of the square d to one of its corners is √ . Using the parallel 2 axis theorem, the moment of inertia of the square about an axis at one of its corners is corner Isquare 2 1 d 2 = msquare d + msquare √ 3 2 1 1 = msquare d2 + 3 2 5 = msquare d2 6 5 = (5 kg) (6 m)2 6 = 150 kg · m2 . 016 10.0 points Determine the moment of inertia of a cylinder of radius 0.71 m, height 2 m and density (1.18 − 0.59 r + 0.339 r 2 ) kg/m3 about the center. Correct answer: 0.765446 kg · m2 . 5 Explanation: Let : I= Z Z r = 0.71 m h = 2 m. and ρ r 2 dV R (1.18 − 0.59 r + 0.339 r 2 ) r 2 r dr 0 Z 2π Z h × dθ dz 0 0 R5 R6 R4 2πh − 0.59 + 0.339 = 1.18 4 5 6 (0.71 m)4 (0.71 m)5 = 1.18 − 0.59 4 5 (0.71 m)6 2 π (2 m) + 0.339 6 = = 0.765446 kg · m2 . 017 10.0 points The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.5 rev/s in 7.7 s . At this point the person doing the laundry opens the lid, and a safety switch turns oﬀ the washer. The tub slows to rest in 13.7 s . During the whole process of starting up and stopping, through how many revolutions does the tub turn? Assume constant angular acceleration while it is starting and stopping. Correct answer: 58.85 rev. Explanation: Let’s split the problem into two pieces, the ﬁrst phase as the washer is starting to spin, and the second phase while it is slowing down. In both cases, we will use our two equations of angular motion: ωf = ω0 + α t 1 θ f = θ 0 + ω0 t + α t2 . 2 Now, during the ﬁrst phase of the motion (when the washer is starting up), its initial angular velocity is zero, ω0 = 0, its ﬁnal angular xu (rx575) – HW08 – macdonald – (55475) velocity is 5.5 rev/s, and the time it takes to reach this angular velocity is 7.7 s. We can solve for the angular acceleration using the ﬁrst equation above: ωf = ω0 + α t 5.5 rev/s = α(7.7 s) 5.5 rev/s ⇒α= = 0.714286 rev/s2 . 7.7 s Now we can use the angular position equation to determine the number of revolutions during the ﬁrst phase of the motion, letting θ0 = 0: 1 θ f = θ 0 + ω0 t + α t2 2 1 θf = (0.714286 rev/s2 )(7.7 s)2 2 ⇒ θf = 21.175 rev . Now we repeat the process for the second phase of the motion, where now the initial angular velocity is ω0 = 5.5 rev/s and the ﬁnal angular velocity is ωf = 0. The time it takes to decelerate is t = 13.7 s. We again solve for the angular deceleration using the angular velocity equation: ωf = ω0 + α t 0 = 5.5 rev/s + α(13.7 s) 5.5 rev/s ⇒α=− = −0.40146 rev/s2 . 13.7 s Now we again use the angular position equation to determine the number of revolutions during the second phase of the motion, letting θ0 = 21.175 rev: 1 θ f = θ 0 + ω0 t + α t2 2 θf = 21.175 rev + 5.5 rev/s(13.7 s) 1 − (0.40146 rev/s2 )(13.7 s)2 2 ⇒ θf = 58.85 rev . Since we included the number of revolutions made during the ﬁrst phase as θ0 in our calculation for the second phase, this is the total number of revolutions. 6 018 (part 1 of 3) 10.0 points What is the tangential acceleration of a bug on the rim of a 78 rpm record of diameter 14.42 in. if the record moves from rest to its ﬁnal angular speed in 4 s? The conversion between inches and meters is 0.0254 m/in. Correct answer: 0.373966 m/s2 . Explanation: Let : ω = 78 rpm , r = (7.21 in) (0.0254 m/in) = 0.183134 m t = 4 s. ω = ω0 + α t = α t ∆ω α= , so ∆t r ∆ω ∆t (0.183134 m) (78 rpm) 2 π 1 min = 4s 1 rev 60 s at = r α = = 0.373966 m/s2 . 019 (part 2 of 3) 10.0 points When the record is at its ﬁnal speed, what is the tangential velocity of the bug? Correct answer: 1.49586 m/s. Explanation: vt = r ω = (0.183134 m)(78 rpm) 2 π 1 min 1 rev 60 s = 1.49586 m/s . 020 (part 3 of 3) 10.0 points What is its radial acceleration 2.6 s after starting from rest? Assume that the record has constant angular acceleration. xu (rx575) – HW08 – macdonald – (55475) Each spoke contributes to the total moment of inertia as a thin rod pivoted at one end, so 2 Correct answer: 5.16229 m/s . Explanation: Since the record has constant angular acceleration, and starts from rest we know that ωf ω = αt = t. tf Knowing this, we know that the relationship between radial acceleration and the velocity of rotation is given by: 2 ωf v2 2 ar = = ω r = t r r tf 2 78 rpm 2 π 0.183134 m = (2.6 s) s 4 s 60 min 2 = 5.16229 m/s . m R2 I = Ihoop + n Irod = M R2 + n 3 m 2 R = M +n 3 0.087 kg = 1 kg + 7 (0.62 m)2 3 = 0.462433 kg · m2 . 022 (part 2 of 2) 10.0 points Determine the moment of inertia of the wheel about an axis through its rim and perpendicular to the plane of the wheel. Correct answer: 1.08093 kg · m2 . Explanation: From the parallel axis theorem 021 (part 1 of 2) 10.0 points A wheel is formed from a hoop of mass 1 kg and seven equally spaced spokes, each of mass 0.087 kg. The hoop’s radius is the length 0.62 m of each spoke. 0.62 m 0.087 kg Irim = I + (M + n m) R2 = 0.462433 kg · m2 + [1 kg + 7 (0.087 kg)] (0.62 m)2 = 1.08093 kg · m2 . 023 10.0 points Find the moment of inertia of a solid sphere of mass M and radius R about an axis that is tangent to the sphere. m 1 kg Find the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel. Correct answer: 0.462433 kg · m2 . Explanation: Let : 7 n = 7, m = 0.087 kg , M = 1 kg , and R = 0.62 m . ω 8 M R2 5 4 2. I = M R2 5 3 3. I = M R2 5 9 4. I = M R2 5 1. I = r xu (rx575) – HW08 – macdonald – (55475) 7 M R2 correct 5 6 6. I = M R2 5 Explanation: The moment of inertia of a solid about a diameter is 8 N 5. I = Icm = θ sin g m P θ mg 2 M R2 . 5 Using the parallel-axis theorem, the moment of inertia about an axis that is tangent to the sphere is The moment of inertia of the ball about the point of contact between the ball and the inclined plane is I = Icm + M R2 2 = M R2 + M R2 5 7 = M R2 . 5 IP = Icm + m d2 2 = m R2 + m R2 3 5 = m R2 . 3 024 10.0 points A 140 g basketball has a 37.7 cm diameter and may be approximated as a thin spherical shell. The net torque about the point of contact between the ball and the inclined plane is 5. 3 9m 5 m R2 α 3 3 m g R2 sin θ = g sin θ . a= 5 5 m R2 3 m g R sin θ = IP α = 140 g 0 µ= . 37 25◦ Starting from rest, how long will it take a basketball to roll without slipping 5.39 m down an incline that makes an angle of 25 ◦ with the horizontal? The moment of inertia of a thin spherical shell of radius R and mass m 2 is I = m R2 , the acceleration due to gravity 3 is 9.8 m/s2 , and the coeﬃcient of friction is 0.37 . Correct answer: 2.08279 s. Explanation: Let : ℓ = 5.39 m , θ = 25◦ , and R = 0.1885 m . Because the sphere starts from rest, its center of mass moves a distance 1 2 at 2 s r 2ℓ 2 (5.39 m) = t= a 2.485 m/s2 ℓ= = 2.08279 s . 025 10.0 points Consider a wheel (solid disk) of radius 1.41 m, 1 mass 15 kg and moment of inertia M R2 . 2 The wheel rolls without slipping in a straight line in an uphill direction 16◦ above the horizontal. The wheel starts at angular speed 14.7518 rad/s but the rotation slows down xu (rx575) – HW08 – macdonald – (55475) as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops? The acceleration of gravity is 9.8 m/s2 . Correct answer: 120.122 m. 9 3 R2 ω02 L= 4 g sin θ 3 (1.41 m)2 (14.7518 rad/s)2 = 4 (9.8 m/s2 ) sin 16◦ = 120.122 m . Explanation: Let : R = 1.41 m , M = 15 kg , θ = 16◦ , ω0 = 14.7518 rad/s , g = 9.8 m/s2 . The actual value of the wheel’s mass is extraneous to this situation. and Since the wheel rolls without slipping, v = R ω the friction force does no work so the net mechanical energy of the wheel is conserved: potential kinetic kinetic Elinear + Erotation + Egravity = const 1 1 M v 2 + I ω 2 + M g y = const 2 2 1 1 1 2 2 M (R ω) + M R ω2 + M g y 2 2 2 = const 3 M R2 ω 2 + M g y = const . 4 The wheel starts at initial angular speed ω = ω0 , but as it rolls uphill, its potential energy increases while the kinetic energy decreases, and eventually the kinetic energy is completely exhausted and the wheel stops. By energy conservation, this happens when 1 2 ω + M g y0 = 0 + M g ystop . 4 0 During all this time, the wheel’s axis moves some distance L in a straight line in the direction θ = 16 ◦ above the horizontal, so the wheel’s elevation y increases by ∆y ≡ ystop − y0 = L sin θ . Therefore, 3 M R2 ω02 = M g ∆y = M g L sin θ . 4 026 10.0 points A uniform solid sphere with moment of inertia 2 m r 2 . mass 1.3 kg and radius 0.378 m is 5 placed on the inside surface of a hemispherical bowl of radius 1.7 m. The sphere is released from rest at an angle 69.2◦ from the vertical and rolls without slipping. R θ r What is the angular velocity of the sphere when it reaches the bottom of the bowl? The acceleration of gravity is 9.8 m/s2 . Correct answer: 9.1397 rad/s. Explanation: The distance between the center of mass of the sphere at its highest (rest) and its lowest points is h = (R − r) − (R − r) cos θ = (R − r)(1 − cos θ) = (1.7 m − 0.378 m)(1 − cos 69.2◦ ) = 0.852549 m . v = r ω , so from conservation of energy. Ui = K f 1 m g h = m vf2 + 2 1 = m vf2 + 2 1 I ω2 2 f vf 2 1 2 2 mr 2 5 r xu (rx575) – HW08 – macdonald – (55475) 7 m vf2 10 r 10 vf = gh 7 = Thus, vf 1 = ωf = r r r 10 gh 7 = 9.1397 rad/s . keywords: 10