PROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION

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Ruth Poole
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1 PROLEM The bent rod E rotates about a line joining Points and E with a constant angular elocity of 9 rad/s. Knowing that the rotation is clockwise as iewed from E, determine the elocity and acceleration of corner E = E = 0.6 m re / = (0.4 m) i+ (0.15 m) j E = (0.4 m) i+ (0.4 m) j+ (0.2 m) k E 1 1 λ E = = ( 0.4i+ 0.4j+ 0.2 k) = ( 2i+ 2 j+ k) E ω= ωeλe = (9 rad/s) ( 2i+ 2 j+ k) 3 ω = (6 rad/s) i+ (6 rad/s) j+ (3 rad/s) k i j k = ω r / = = 0.45i 1.2 j+ ( ) k E a = a r + ω ( ω r ) = a r + ω a E / E / E / i j k = = ( ) i+ ( ) j+ ( ) k = (0.45 m/s) i (1.2 m/s) j+ (1.5 m/s) k a = (12.60 m/s ) i+ (7.65 m/s ) j+ (9.90 m/s ) k

PROLEM 15.12 The rectangular block shown rotates about the diagonal O with a constant angular elocity of 6.76 rad/s.

2 PROLEM The rectangular block shown rotates about the diagonal O with a constant angular elocity of 6.76 rad/s. Knowing that the rotation is counterclockwise as iewed from, determine the elocity and acceleration of point at the instant shown. Gien: ω = 6.76 rad/s ( ) ( ) ( ) r O / = 5 in. i in. j + 12 in. k ( ) ( ) r O / = 5 in. i in. j ( ) ( ) ( ) l = = 33.8 in. O ngular elocity: Velocity of point : cceleration of point : a ω 6.76 ω = ro / = 5 i j + 12 k 33.8 l O ( ) ( 1.0 rad/s) ( 6.24 rad/s) ( 2.4 rad/s) ω = i + j + k = ω r O / i j k = = 37.44i + 12j 15.6k = ω ( 37.4 in./s) ( in./s) ( in./s) = i + j k i j k a = = 126.1i 74.26j 245.6k ( in./s 2) ( 74.3 in./s 2 2 ) ( 246 in./s ) a = i j + k

PROLEM 15.22 The two pulleys shown may be operated with the V belt in any of three positions.

3 PROLEM The two pulleys shown may be operated with the V belt in any of three positions. If the angular acceleration of shaft is 6 rad/s 2 and if the system is initially at rest, determine the time required for shaft to reach a speed of 400 rpm with the belt in each of the three positions. ngular elocity of shaft : ω t elt speed: = rω = rω rα t ngular speed of shaft : ω = = r r rω Soling for t, t = r α ata: a = 6 rad/s, ω = 400 rpm = rad/s elt at left: elt in middle: elt at right: r r d t = = = r 6 r d d 2 in. d = 4 in. t = 3.49 s d 3 in. d = 3 in. t = 6.98 s d 4 in. d = 2 in. t = s

PROLEM 15.24 gear reduction system consists of three gears,, and.

4 PROLEM gear reduction system consists of three gears,, and. Knowing that gear rotates clockwise with a constant angular elocity ω = 600 rpm, determine (a) the angular elocities of gears and, (b) the accelerations of the points on gears and which are in contact. (a) (600)(2 p ) ω = 600 rpm = = 20p rad/s. 60 Let Points,, and lie at the axles of gears,, and, respectiely. Let be the contact point between gears and. = r / ω= (2)(20 π) = 40π in./s 40p 60 ω = = = 10p rad/s = 10p = 300 rpm r 4 2p / Let E be the contact point between gears and. E = re / ω = (2)(10 π) = 20π in./s ω = 300 rpm E 20p 60 ω = = = 3.333p rad/s = (3.333 p) = 100 rpm r 6 2p E / (b) ccelerations at Point E. ω = 100 rpm On gear : a 2 2 E (20 π ) = = = in./s r 2 E / 2 a = 1974 in./s E (20 π ) On gear : a = = = 658 in./s r 6 E / 2 a = 658 in./s 2

PROLEM 15.40 painter is halfway up a 10-m ladder when the bottom starts sliding out from under him.

Gien: = 2 i m/s, θ =60 Geometry: r = 10 cos 60 + i 10sin 60 j Relatie Velocity: Equate omponents: / r = 5cos 60 + i 5sin 60 j P/ = + = + ω k r / / 2 ω ( 5 8.

5 PROLEM painter is halfway up a 10-m ladder when the bottom starts sliding out from under him. Knowing that point has a elocity = 2 m/s directed to the left when θ= 60, determine (a) the angular elocity of the ladder, (b) the elocity of the painter. Gien: = 2 i m/s, θ =60 Geometry: r = 10 cos 60 + i 10sin 60 j Relatie Velocity: Equate omponents: / r = 5cos 60 + i 5sin 60 j P/ = + = + ω k r / / 2 ω ( ) ( ω ) i 5ω j j= i+ k i+ j = + + (a) i: 0 = ( ω ) ω = rad/s ω = rad/s Velocity of Painter (point P): (b) = + = + ω k r P P / P/ ( ) = 2i 0.231k 2.5i j ( 1.00 m/s) ( m/s) P = i j

PROLEM 15.57 Knowing that the disk has a constant angular elocity of 15 rad/s clockwise, determine the angular elocity of bar and the elocity of collar when (a) θ = 0, (b) θ = 90, (c) θ = 180.

6 PROLEM Knowing that the disk has a constant angular elocity of 15 rad/s clockwise, determine the angular elocity of bar and the elocity of collar when (a) θ = 0, (b) θ = 90, (c) θ = 180. isk ar : Rotation about a fixed axis. ω = 15 rad/s, = 2.8 in. ( ) ω ( )( ) = = = 42 in./s ( a) θ = in./s = 2.8 sin β =, β = = + / = [42 ] + / β / 42 = = in./s cos β / ω = =, 4.38 rad/s 10 ω = ( b) θ = in./s = tan β, in./s = 5.6 sin β =, β = ar : = + / = = [42 ] + / β omponents: : / = 0 : = ω 0 = 42.0 in./s =

7 PROLEM (ontinued) () c θ = 180. = 42 in./s 2.8 sin β =, β = ar : = + / = [42 ] + / β / 42 = = cos β / ω = = 10 ω 4.38 rad/s = = tan β in./s =

PROLEM 15.61 In the engine system shown, l = 160 mm and b = 60 mm.

8 PROLEM In the engine system shown, l = 160 mm and b = 60 mm. Knowing that the crank rotates with a constant angular elocity of 1000 rpm clockwise, determine the elocity of the piston P and the angular elocity of the connecting rod when (a) θ = 0, (b) θ = 90. ω = 1000 rpm (1000)(2 π ) = = rad/s 60 (a) θ = 0. rank. (Rotation about ) r / 0.06 m = Rod. (Plane motion = Translation with + Rotation about ) = + / = [ ] + [ / ] / P = 0 = m/s (b) θ = 90. rank. (Rotation about ) r / 0.06 m = ω = = 39.3 rad/s l 0.16 ω = = = r / ω = (0.06)(104.72) = m/s Rod. (Plane motion = Translation with + Rotation about.) = + / [ ] ] = [6.2832] + [ / β ] / = 0, = m/s / ω = ω = 0 l = = m/s 6.28 m/s P = r / ω = (0.06)(104.72) = m/s P = P =

9 PROLEM In the position shown, bar E has a constant angular elocity of 10 rad/s clockwise. Knowing that h = 500 mm, determine (a) the angular elocity of bar F, (b) the elocity of Point F. ar E: (Rotation about E) ω E = 10 rad/s = (10 rad/s) k r E / = (0.1 m) i+ (0.2 m) j = ωe r/ E = ( 10 k) ( 0.1i+ 0.2 j) = (1 m/s) j+ (2 m/s) i ar F: (Plane motion = Translation with + Rotation about.) ar : (Rotation about ) Equating components of the two expressions for, ω = ωk r/ = (0.3 m) i+ (0.1 m) j = + ω r/ = j+ 2 i+ ( ωk) ( 0.3i+ 0.1 j) = j+ 2i 0.3ω j 0.1ω i ω = ωk r/ = (0.42 m) j = ω r/ = ( ωk) (0.42 j) = 0.42ωi (a) j : 1 0.3ω = 0 ω = rad/s ω 3.33 rad/s i : 2 0.1ω = 0.42ω 2 (0.1)(3.3333) = 0.42ω ω = rad/s ω = 3.97 rad/s h ar F: rf / = r / where = 0.3 F = + ω rf/ = j+ 2 i+ ( 0.3ωj 0.1 ωi) = j+ 2 i+ ( j i) With 0.8 h= 500 mm = 0.5 m, = = F = j+ 2 i j i (b) ( m/s) i ( m/s) j 2.00 m/s F = = F = 56.3

PROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION. ω = 29.6 rad/s. ω = = 36 3.

PROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION. ω = 29.6 rad/s. ω = = 36 3. 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