Version 001 Unit 1 Rotational Kinematics baker (BC303) 1 This print-out should have 0 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Angular Speed of a Record 001 (part 1 of ) 10.0 points A record has an angular speed of 5 rev/min. What is its angular speed? Correct answer:.618 rad/s. Let : ω = 5 rev/min. ( )( ) π rad 1 min ω = (5 rev/min) rev 60 s =.618 rad/s. The linear speed is v = Rω ω = v R = 44.9 m/s 363 m = 0.13691 rad/s. 004 (part of ) 10.0 points Find the magnitude of its acceleration. Correct answer: 5.55375 m/s. With the car moving at a constant speed, there is no tangential acceleration, so the acceleration is purely radial: a r = v R = (44.9 m/s) 363 m = 5.55375 m/s. 00 (part of ) 10.0 points Through what angle does it rotate in 1.99 s? Correct answer: 5.0981 rad. Let : t = 1.99 s. θ = ωt = (.618 rad/s)(1.99 s) = 5.0981 rad. Circular Race Track 03 003 (part 1 of ) 10.0 points A racing car travels on a circular track of radius 363 m, moving with a constant linear speed of 44.9 m/s. Find its angular speed. Correct answer: 0.13691 rad/s. Let : v = 44.9 m/s and R = 363 m. Rotating Ladybug 005 10.0 points A ladybug sits at the outer edge of a merrygo-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. What is the gentleman bug s angular speed? 1. twice the ladybug s. half the ladybug s 3. impossible to determine 4. the same as the ladybug s correct Angular speed is the same for every point on the merry-go-round. Grinding Wheel Motor 006 10.0 points An electric motor rotating a workshop grinding wheel at a rate of 147 rev/min is switched off. Assume constant angular deceleration of magnitude 1.78 rad/s.
Version 001 Unit 1 Rotational Kinematics baker (BC303) Through how many revolutions does the wheel turn before it finally comes to rest? Correct answer: 10.5941 rev. Let : ω 0 = 147 rev/min, α = 1.78 rad/s, and ω f = 0. From kinematics, ω f = ω 0 +αθ θ = ω f ω 0 = ω 0 α α (147 rev/min) = ( 1.78 rad/s ) π rad 1 rev = 10.5941 rev. ( ) 1 min 60 s AP M 1993 MC 0 007 10.0 points A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angular velocity of the turntable after it has made one complete revolution? 1. ω = 4πα. ω = α 3. ω = πα correct 4. ω = πα 5. ω = α Similar to uniform linear accelerated motion, we have ω f = ω 0 +αθ = α(π) = 4πα ω f = πα Spinning Disk 01 008 10.0 points A disk is initially spinning about a fixed axis and slowing down with constant angular acceleration α. It comes to rest in time t. Through what angle does it turn before coming to rest? 1. Since ω(0) is not given, the problem has no solution.. θ = α ω(0) t 3. θ = ω(0)t+ 1 αt 4. θ = [ω(0)] α 5. θ = 1 αt correct 6. None of these 7. θ = αt 8. θ = ω(0)t αt 9. θ = αt The disk is at rest after time t, so ω = ω(0) αt = 0, ω(0) = αt and 6. ω = πα Let : ω 0 = 0 rad/s and θ = 1 rev = π rad. θ = ω(0)t 1 αt = αt 1 αt = 1 αt
Version 001 Unit 1 Rotational Kinematics baker (BC303) 3 Decelerating a Wheel 009 10.0 points Initiallya wheel rotatingabout afixed axisat a constant angular deceleration of 0.6 rad/s has an angular velocity of 0 rad/s and an angular position of 9.1 rad. What is the angular position of the wheel after 3.3 s? Correct answer: 5.833 rad. Let : α = 0.6 rad/s, ω 0 = 0 rad/s, θ 0 = 9.1 rad, and t = 3.3 s. From kinematics, θ = ω 0 t+ 1 αt +θ 0 = (0 rad/s)(3.3 s) + 1 ( 0.6 rad/s )(3.3 s) +9.1 rad = 5.833 rad. Concept 08 0 010 10.0 points A large wheel is coupled to a wheel with half the diameter as shown. r How does the rotational speed of the smaller wheel compare with that of the larger wheel? How do the tangential speeds at the rims compare (assuming the belt doesn t slip)? 1. The smaller wheel has twice the rotational r speed and twice the tangential speed as the larger wheel.. The smaller wheel has half the rotational speed and half the tangential speed as the larger wheel. 3. The smaller wheel has twice the rotational speed and the same tangential speed as the larger wheel. correct 4. The smaller wheel has four times the rotational speed and the same tangential speed as the larger wheel. v = rω The tangential speeds are equal, since the rims arein contact withthe belt and have the same linear speed as the belt. The smaller wheel (with half the radius) rotates twice as fast: ( ) 1 r (ω) = rω = v Concept 08 03 011 10.0 points An automobile speedometer is configured to read speed proportional to the rotational speed of its wheels. If larger wheels, such as those of snow tires, are used, what will be the effect on the speedometer reading? 1. The speedometer reads a higher speed than you are moving. The speedometer shows correctly the speed you are moving 3. The speedometer reads a lower speed than you are moving correct 4. The sign of the corrected speedometer speed is speed dependent (A speedometer actually meansures the RPM of the wheels and displays this as mi/h
Version 001 Unit 1 Rotational Kinematics baker (BC303) 4 or km/h. The conversion from RPM to the mi/h or km/h reading assumes the wheels are a certain size.) Oversized wheels give too low a reading, because they really travel farther per revolution than the speedometer indicates; undersized wheels give too high a reading because the wheels do not go as far per revolution. keywords: Tipler PSE5 09 47 01 10.0 points A 1.4 m diameter wagon wheel consists of a thinrim having a mass of8kg andsix spokes, each with a mass of 0.9 kg. = M rim R +M spoke L = (8 kg)(0.7 m) +(0.9 kg)(0.7 m) = 4.80 kg m. Rod and Disc 013 10.0 points Consider the rigid rod-plus-disc system P 1.4 m 0.9 kg 8 kg Find the moment of inertia of the wagon wheel for rotation about its axis. Correct answer: 4.80 kg m. Let : L = R = 0.7 m, M rim = 8 kg, and M spoke = 0.9 kg. The moment of inertia of the wagon wheel comes from two parts: the rim and the spokes Therefore I rim = M rim R, I spokes = 1 3 M spokel. I wheel = M rim R +6 ( ) 1 3 M spokel Findthemomentofinertiaofthependulum as it freely rotates about the point P. The rod has a length 1.39 m, and the disc has a radius halfthat. ThepivotatPisafourthoftheway from the end of the rod. The rod has a mass of 0.64 kg and the disc is twice as massive with uniform mass distribution. Correct answer: 1.88058 kg m. Let : l = 1.39 m, m = 0.64 kg, r = 0.695 m, and m d = 1.8 kg. Apply the parallel-axis theorem to both the rod and the disc: and I rod = 1 3 m ( L ) +m ( ) L = 7 4 48 ml I disc = 1 ( 3R (m)r +(m) = 11 mr, so )
Version 001 Unit 1 Rotational Kinematics baker (BC303) 5 I tot = I rod +I disc = 7 48 ml + 11 mr = 7 (0.64 kg)(1.39 m) 48 + 11 (0.64 kg)(0.695 m) = 1.88058 kg m. Inertia of Disk Ring Square 014 10.0 points Consider three objects of equal masses but different shapes: a solid disk, a thin ring, and athin hollowsquare. The ring and thesquare are hollow and their perimeters carry all the mass, but the disk is solid and has uniform mass density over its whole area. disk ring square For the square the distance from the center ranges from R to R, so and Thus r > R I cm disk < M R I cm ring = M R I cm square > M R. I cm square > Icm ring > Icm disk Tipler PSE5 09 46 015 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. R R R Compare the three objects moments of inertia when rotated around their respective centers of mass. 1. I cm ring > Icm disk > Icm square m ω r. I cm disk > Icm ring > Icm square 3. I cm square > I cm ring > I cm disk correct 4. I cm square > Icm disk > Icm ring 5. I cm disk > Icm square > I cm ring 6. Iring cm > Icm square > Icm disk The moment of inertia I = r dm is I = M r, where M isthe object s net mass and r is the average distance of massive points in the object from the rotational axis. For a hoop or a thin ring, all massive points are at the same distance R from the axis, so r = R 1. I = 8 5 M R. I = 9 5 M R 3. I = 7 5 M R correct 4. I = 4 5 M R 5. I = 3 5 M R 6. I = 6 5 M R The moment of inertia of a solid about a diameter is I cm = 5 M R.
Version 001 Unit 1 Rotational Kinematics baker (BC303) 6 Using the parallel-axis theorem, the moment of inertia about an axis that is tangent to the sphere is I = I cm +M R = 5 M R +M R 3.3 m 5.5 kg 1.5 m = 7 5 M R. SWCT Inertia 016 10.0 points A particular flywheel that rotates about its center of mass has a moment of inertia I = 3 4 MR, where R = the radius. What is the moment of inertia if the flywheel is rotated about a point on its rim? 1. 7 4 MR correct. 3 4 MR 3. 3MR 4. 5 4 MR 5. 3 MR I = MR + 3 4 MR = 7 4 MR Grandfather Clock Pendulum 017 10.0 points A pendulum is made of a rod of mass 5.5 kg and length 3.3 m whose moment of inertia about its center of mass is 1 1 M L and a thin cylindrical disk of mass.1 kg and radius 1.5 m whose moment of inertia about its center of mass is 1 M R..1 kg ω What is the moment of inertia of the pendulum about the pivot point? Correct answer: 45.1965 kg m. Let : L = 3.3 m, M r = 5.5 kg, R = 1.5 m, and M d =.1 kg. By the parallel axis theorem I pp = I cm +M L, where I cm is the moment of inertia about the center of mass, M is the mass, and L is the distance from the center of mass to the pivot point. l = L = 1.65 m, I rod = 1 1 M rl +M r l, I disk = 1 M dr +M d L, and the total moment of inertia is I = I rod +I disk = 1 1 M rl +M r l + 1 M dr +M d L = 1 (5.5 kg)(3.3 m) 1 +(5.5 kg) (1.65 m) + 1 (.1 kg)(1.5 m) +(.1 kg)(3.3 m) = 45.1965 kg m.
Version 001 Unit 1 Rotational Kinematics baker (BC303) 7 Moment of Inertia of a Rod 018 10.0 points The moment of inertia about the center of mass of a rod of length L is I = ml /1. Its moment of inertia about its end point is 3. ml 4. 4ml 5. 1 ml 1. ml 3. ml 4 3. ml correct 6. ml 7. 3ml 8. 6ml 4. ml 5. ml 6 By the parallel-axis theorem where I = I CM +md, d = L is the distance from the center-of-mass axis to the parallel axis, we have I = ml 1 + ml 4 = ml 3. The center of mass is at the geometrical center of the system, as is easily seen by symmetry, or by using r cm = 1 r i projected 4 i along the x axis. Let the center of mass be at x = 0; then I cm = [ (3 m i x i ) = ml + i = 5ml. ( ) ] 1 Four Point Masses 019 10.0 points Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible mass. The distance between any two adjacent masses is l. l l l m m m m What is the rotational inertia I cm of this system about its center of mass? 1. 5ml correct. 7ml Hinged Rod With Mass 07 00 10.0 points Consider a thin rod of length L which is pivoted at one end. A uniformly dense spherical object (with mass m and radius r = 1 ) 8 L is attached to the free end of the rod. The moment of inertia of the rod about an end is I rod = 1 3 ml and the moment of inertia of the sphere about its center of mass is I sphere = 5 mr.
Version 001 Unit 1 Rotational Kinematics baker (BC303) 8 = 1 ( ) 9 3 ml + ml + 1 8 160 ml = 1541 960 ml. C L m θ r= 1 8 L m Find the moment of inertia of the rod plus mass system with respect to the pivot point. The acceleration of gravity is 9.8 m/s. 1. I = 637 405 ml. None of these 3. I = 1541 960 ml correct 4. I = 173 105 ml 5. I = 307 180 ml 6. I = 671 375 ml Rotational kinetic energy is K rot = 1 Iω. Themoment ofinertiaoftherodwithrespect to the pivot point is I rod = 1 3 ml, and the moment of inertia of the mass m with respect to the pivot point is I sphere = 5 mr +m(l+r), so the moment of inertia of the system is I = I rod +I m = 1 3 ml +m(l+r) + 5 mr = 1 ( 3 ml + 1+ 1 ) ml 8 + 5 m ( 1 8 L )