PROLEM 15.1 The brake drum is attached to a larger flywheel that is not shown. The motion of the brake drum is defined by the relation θ = 36t 1.6 t, where θ is expressed in radians and t in seconds. Determine (a) the angular elocity at t = s, (b) the number of reolutions executed by the brake drum before coming to rest. Gien: Differentiate to obtain the angular elocity. θ = 36t 1.6t radians dθ ω = = 36 3.t rad/s dt (a) t t = s, ω = 36 (3.)() ω = 9.6 rad/s (b) When the rotor stops, ω = 0. 0 = 36 3.t t = 11.5 s θ = (36)(11.5) (1.6)(11.5) = 0.5 radians In reolutions, 0.5 θ = θ = 3. re π
PROLEM 15.7 When studying whiplash resulting from rear end collisions, the rotation of the head is of primary interest. n impact test was performed, and it was found that the angular acceleration of the head is defined by the relation α = 700 cosθ + 70sinθ where α is expressed in rad/s and θ in radians. Knowing that the head is initially at rest, determine the angular elocity of the head when θ = 30. ngular motion relations: Separating ariables ω and θ gies dω ωdω α = = = 700 cosθ + 70sinθ dt dθ ωdω = (700 cosθ + 70sin θ) dθ Integrating, using ω = 0 when θ = 0, ω θ ωdω = (700 cosθ + 70sin θ) dθ 0 0 π Data: θ = 30 = rad 6 1 ω = (700sinθ 70 cos θ) = 700sinθ + 70(1 cos θ) ω = 1400sinθ + 140(1 cos θ) With calculator set to degrees for trigonometric functions, ω = 1400sin 30 + 140(1 cos 30 ) = 6.8 rad/s θ 0 ω = 6.8 rad/s With calculator set to radians for trigonometric functions, ω = 1400sin( π/6) + 140(1 cos( π/6)) = 6.8 rad/s
PROLEM 15.8 The angular acceleration of an oscillating disk is defined by the relation α = kθ. Determine (a) the alue of k for which ω = 1 rad/s when θ = 0 and θ = 6 rad when ω = 0, (b) the angular elocity of the disk when θ = 3 rad. Gien: a k and = d ω = θ a ω d θ ω dω = αdθ ω dω = kθdθ Integrating, 0 6 = 1 0 ω dω k θdθ 1 6 0 = k 0 ( a ) k 1 = = 9s 6 k = 9.00 s 3 = 1 0 ω ω dω k θdθ () b ( )( ) ω 1 3 = 9 0 ω = 1 9 3 = 63 rad /s ω = 7.94 rad/s
PROLEM 15.18 series of small machine components being moed by a coneyor belt pass oer a 10 mm radius idler pulley. t the instant shown, the elocity of Point is 300 mm/s to the left and its acceleration is 180 mm/s to the right. Determine (a) the angular elocity and angular acceleration of the idler pulley, (b) the total acceleration of the machine component at. = = 300 mm/s r = 10 mm (a) = ωr, (b) ( a ) = ar, t ( a ) = a = 180 mm/s t 300 ω = = =.5 rad/s ω =.50 rad/s r 10 ( a) t 180 a = = = 1.5 rad/s r 10 ( a ) = r ω = (10)(.5) = 750 mm/s n a = 1.500 rad/s = ( ) t+ ( ) n= (180) + (750) = 771 mm/s a a a 750 tan β =, β = 76.5 180 a = 771 mm/s 76.5
PROLEM 15.8 plastic film moes oer two drums. During a 4-s interal the speed of the tape is increased uniformly from 0 = ft/s to 1 = 4 ft/s. Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum, (b) the number of reolutions executed by drum during the 4-s interal. elt motion: = 0 + at 4 ft/s = ft/s + a(4 s) 4 ft/s ft/s a = = 0.5 ft/s = 6 in./s 4s Since the belt does not slip relatie to the periphery of the drum, the tangential acceleration at the periphery of the drum is (a) ngular acceleration of drum. a = 6 in./s t (b) ngular displacement of drum. at a = = r 6 in./s 15 in. 0 4 in./s t t = 0, ω 0 = = = 1.6 rad/s r 15 in. 1 48 in./s t t = 4 s, ω 1 = = = 3. rad/s r 15 in. α = 0.400 rad/s In reolutions, 1 0 ω = ω + aθ 1 ω0 (3.) (1.6) ω θ = = = 9.6 radians a ()(0.400) θ = 9.6 θ = π 1.58 re
PROLEM 15.31 load is to be raised 0 ft by the hoisting system shown. ssuming gear is initially at rest, accelerates uniformly to a speed of 10 rpm in 5 s, and then maintains a constant speed of 10 rpm, determine (a) the number of reolutions executed by gear in raising the load, (b) the time required to raise the load. The load is raised a distance h = 0 ft = 40 in. For gear-pulley, radius to rope grooe is r 1 = 15 in. h 40 Required angle change for : θ = = = 16 radians r1 15 Circumferential trael of gears and : s = rθ = r θ where r = 18 in. and r = 3 in. (a) ngle change of gear : s = (18 in.)(16 radians) = 88 in. s 88 θ = = = 96 radians r 3 96 In reolutions, θ = θ = 15.8 re π (b) Motion of gear. ω0 = 0, ω = 10 rpm = 4p rad/s Gear is uniformly accelerated oer the first 5 seconds. f ωf ω 4π rad/s a = = =.5133 rad/s t 5s 0 1 1 (.5133)(5) t 31.416 radians θ = a = = The angle change oer the constant speed phase is For uniform motion, θ = ω f ( t) θ = θ θ = 96 31.416 = 64.584 radians θ 64.584 t = = = 5.139 s ω 4π f Total time elapsed: tf = 5s+ t t = 10.14 s f
PROLEM 15.38 n automobile traels to the right at a constant speed of 48 mi/h. If the diameter of a wheel is in., determine the elocities of Points, C, D, and E on the rim of the wheel. = 48 mi/h = 70.4 ft/s C = 0 d d = in. r = = 11 in. = 0.91667 ft ω = 70.4 76.8 rad/s r = 0.91667 = = = = rω / D / E / = (0.91667)(76.8) = 70.4 ft/s = + / = [70.4 ft/s ] + [70.4 ft/s ] D = + D / = [70.4 ft/s ] + [70.4 ft/s 30 ] D = 140.8 ft/s = 136.0 ft/s 15.0 E = + E / = [70.4 ft/s ] + [70.4 ft/s ] E = 99.6 ft/s 45.0
PROLEM 15.41 Rod can slide freely along the floor and the inclined plane. t the instant shown the elocity of end is 1.4 m/s to the left. Determine (a) the angular elocity of the rod, (b) the elocity of end of the rod. Geometry: Velocity analysis: 0.3 sin β =, 0.5 β = 36.87 0.3 tan θ =, 0.15 θ = 67.38 = 1.4 m/s / = rω = 0.5ω β = θ Plane motion = Translation with + Rotation about. Draw elocity ector diagram. Law of sines: = + / ( ) ϕ = 180 θ 90 β = 59.49 / = = sinθ sin 90 β sinϕ / ( ) sinθ 1.4sin 67.38 = = = 1.5 m/s sinϕ sin 59.49 ( a ) () b 1.5 ω = = 3.00 rad/s ω = 3.00 rad/s 0.5 cos β 1.4 cos 36.87 = = = 1.3 m/s sinϕ sin 59.49 = 1.300 m/s 67.4