A6 Solution 2009
PROBLEM 15.158 Four pins slide in four separate slots cut in a circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. If each pin maintains the same velocity relative to the plate when the plate rotates about O with a constant counterclockwise angular velocity ω , determine the acceleration of each pin.
SOLUTION a P = a P′ + a P/F + ac
For each pin: Acceleration of the coinciding point P′ of the plate. For each pin, a P′ = rω 2 towards the center O. Acceleration of the pin relative to the plate.
For pins P1, P2 , and P4 ,
a P/F = 0
For pin P3 ,
a P/F =
u2 r
Coriolis acceleration ac . For each pin ac = 2ω u with ac in a direction obtained by rotating u through 90° in the sense of ω, i.e. . Then,
a1 = rω 2 a 2 = rω 2 a3 = rω 2 a 4 = rω 2
+ [ 2ω u + [ 2ω u u2 + r
+ [ 2ω u
]
a1 = rω 2i − 2ω uj W
]
a 2 = 2ω ui − rω 2 j W + [ 2ω u
]
]
u2 a3 = − rω 2 + + 2ω u i W r
(
)
a 4 = rω 2 + 2ω u j W
PROBLEM 15.188 A 60-mm-radius disk spins at the constant rate ω 2 = 4 rad/s about an axis held by a housing attached to a horizontal rod that rotates at the constant rate ω 1 = 5 rad/s. For the position shown, determine (a) the angular acceleration of the disk, (b) the acceleration of point P on the rim of the disk if θ = 0, (c) the acceleration of point P on the rim of the disk if θ = 90o.
SOLUTION ω = ω1i + ω 2k
Angular velocity.
ω = ( 5 rad/s ) i + ( 4 rad/s ) k (a) Angular acceleration. Frame Oxyz is rotating with angular velocity Ω = ω1i . =ω Oxyz + Ω × ω α =ω
= 0 + ω1i × (ω1i + ω 2k ) = −ω1ω 2 j = − ( 4 )( 5 ) j = −20 j
(
)
) (
)
(
)
α = − 20.0 rad/s 2 j W
(b) θ = 0. Acceleration at point P. rP = ( 60 mm ) i = ( 0.06 m ) i v P = ω × rP = ( 5i + 4k ) × 0.06i = 0.24 j a P = α × rP + ω × v P = −20 j × 0.06i + ( 5i + 4k ) × 0.24 j
= 1.2k + 1.2k − 0.96i = −0.96i + 2.4k
(
a P − 0.960 m/s 2 i + 2.40 m/s 2 k W
(c) θ = 90°. Acceleration at point P. rP = ( 0.06 m ) j v P = ω × rP = ( 5i + 4k ) × 0.06 j = −0.24i + 0.3k a P = α × rP + ω × v P = −20 j × 0.06 j + ( 5i + 4k ) × ( −0.24i + 0.3k ) = 0 + 0 − 1.5j − 0.96 j + 0 = −2.46 j a P = − 2.46 m/s 2 j W
PROBLEM 15.217 Y B
u
360 mm
The bent rod shown rotates at the constant rate ω 1 = 5 rad/s and collar C moves toward point B at a constant relative speed u = 975 mm/s. Knowing that collar C is halfway between points B and D at the instant shown, determine its velocity and acceleration.
C A 150 mm
E
520 mm
w1
X
D
7.
SOLUTION Geometry.
rB = (0.36 m)j,
rD = (0.15 m) i + (0.52 m)k
r B/D = – (0.15 m) i + (0.36 m) j – (0.52 m) k lDB =
>0.15C + >0.36C + >0.52C 2
2
2
= 0.65 m
rC = 1 (rB + rD) = (0.075 m) i + (0.18 m)j + (0.26 m) k 2 9 = w1k = (5 rad/s) k
Let frame Axyz rotate with angular velocity Velocity Analysis.
u = 0.975 m/s vC ¢ = 9 ´ rC = 5k ´ (0.075i + 0.18j + 0.26k) = – (0.9 m/s) i + (0.375 m/s)j
vC/F =
0.975 u rD/B = (– 0.15i + 0.36j – 0.52 k) = – (0.225 m/s)i + (0.54 m/s) j – (0.78 m/s) k 0.65 l DB vC = vC ¢ + vC/F vC = – (1.125 m/s)i + (0.915 m/s) j – (0.78 m/s)k J
Acceleration Analysis.
aC/F = 0
aC ¢ = 9 ´ vC ¢ = 5k ´ (0.9i + 0.375j) = – (1.875 m/s2) i – (4.5 m/s2) j aC/F = 0 29 ´ vC/F = (2) (5k) ´ (0.225i + 0.54j – 0.78k) = – (5.4 m/s2) i – (2.25 m/s2) j aC = aC ¢ + aC/F + 29 ´ vC/F a C = – (7.28 m/s2) i – (6.75 m/s2) j J
PROBLEM 15.237 In the position shown the thin rod moves at a constant speed u = 60 mm/s out of the tube BC. At the same time tube BC rotates at the constant rate ω 2 = 1.5 rad/s with respect to arm CD. Knowing that the entire assembly rotates about the X axis at the constant rate ω 1 = 1.2 rad/s, determine the velocity and acceleration of end A of the rod.
SOLUTION rA = (120 mm ) i + (180 mm ) k
Geometry. Method 1
Let the rigid body DCB be a rotating frame of reference. Its angular velocity is
Ω = ω1i + ω 2k = (1.2 rad/s ) i + (1.5 rad/s ) k.
Its angular acceleration is
α = ω1i × ω 2k = −ω1ω 2 j = − 1.8 rad/s 2 j.
(
)
Motion of the coinciding point A′ in the frame. v A′ = Ω × rA = (1.2i + 1.5k ) × (120i + 180k ) = −216 j + 180 j = − ( 36 mm/s ) j a A′ = α × rA + Ω × v A′ = ( −1.8j) × (120i + 180k ) + (1.2i + 1.5k ) × ( −36 j)
(
) (
)
= 216k − 324i − 43.2k + 54i = − 270 mm/s 2 i + 172.8 mm/s 2 k
Motion of point A relative to the frame. v A/F = ui = ( 60 mm/s ) i,
Velocity of point A.
a A/F = 0
v A = v A′ + v A/F v A = ( 60.0 mm/s ) i − ( 36.0 mm/s ) j W
Coriolis acceleration.
2Ω × v A/F
(
)
2Ω × v A/F = ( 2 )(1.2i + 1.5k ) × 60i = 180 mm/s 2 j
Acceleration of point A.
a A = a A′ + a A/F + 2Ω × v A/F
(
) (
) (
)
a A = − 270 mm/s 2 i + 180.0 mm/s 2 j + 172.8 mm/s 2 k W
PROBLEM 15.235 CONTINUED Method 2 Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity Ω = ω1i = 1.2i rad/s. Then the motion relative to the frame consists of the rotation of body DCB about the z axis with angular velocity ω 2k = (1.5 rad/s ) k plus the sliding motion u = ui = ( 60 mm ) i of the rod AB relative to the body DCB.
Motion of the coinciding point A′ in the frame. v A′ = Ω × rA = 1.2i × (120i + 180k ) = − ( 216 mm/s ) j
(
)
a A′ = Ω × v A′ = 1.2i × ( −216 j) = − 259.2 mm/s 2 k
Motion of point A relative to the frame. v A/F = ω 2k × rA + ui = 1.5k × (120i + 180k ) + 60i = (180 mm/s ) j + ( 60 mm/s ) i a A/F = α 2k × rA + ω 2k × (ω 2k × rA ) + ui + 2ω 2 j × ( ui ) = 0 + 1.5k × 180 j + 0 + ( 2 )(1.5k ) × ( 60i )
(
) (
)
= 180 mm/s 2 j − 270 mm/s 2 i
Velocity of point A.
v A = v A′ + v A/F v A = −216 j + 180 j + 60i v A = ( 60 mm/s ) i − ( 36 mm/s ) j W
Coriolis acceleration.
2Ω × v A/F
(
)
2Ω × v A/F = ( 2 )(1.2i ) × (180 j + 60i ) = 432 mm/s 2 k
Acceleration of point A.
a A = a A′ + a A/F + 2Ω × v A/F a A = −259.2k + 180 j − 270i + 432k
(
) (
) (
)
a A = − 270 mm/s 2 i + 180.0 mm/s 2 j + 172.8 mm/s 2 k W
PROBLEM 15.233 The arm AB of length 5 m is used to provide an elevated platform for construction workers. In the position shown, arm AB is being raised at the constant rate dθ / dt = 0.25 rad/s; simultaneously, the unit is being rotated counterclockwise about the Y axis at the constant rate ω 1 = 0.15 rad/s. Knowing that θ = 20°, determine the velocity and acceleration of point B.
SOLUTION Frame of reference. Let moving frame Axyz rotate about the Y axis with angular velocity Ω = ω1j = ( 0.15 rad/s ) j.
rB/ A = −5cos 20°i + 5sin 20° j = − ( 4.6985 m ) i + (1.7101 m ) j
Geometry.
Place point O on Y axis at same level as point A.
rB/O = rB/ A + rA/O = rB/ A + ( 0.8 m ) i = − ( 3.8985 m ) i + (1.7101 m ) j
Motion of corresponding point B′ in the frame. v B′ = Ω × rB/O = ( 0.15 j) × ( −3.8985i + 1.7101j) = ( 0.58477 m/s ) k
(
)
a B′ = Ω × v B′ = ( 0.15j) × ( 0.58477k ) = 0.087715 m/s 2 i ω2 = −
Motion of point B relative to the frame.
dθ k = ( 0.25 rad/s ) k dt
v B/F = ω2 × rB/ A = ( −0.25 ) k × ( −4.6985i + 1.7101j) = − ( 0.42753 m/s ) i + (1.17462 m/s ) j
a B/F = ω2 × v B/F = ( −0.25k ) × ( −0.42753i + 1.17462 j)
(
) (
)
= 0.29365 m/s 2 i + 0.106881 m/s 2 j
Velocity of point B.
v B = v B′ + v B/F v B = − ( 0.428 m/s ) i + (1.175 m/s ) j + ( 0.585 m/s ) k W
Coriolis acceleration.
2Ω × v B/F = ( 2 )( −0.15j) × ( −0.42753i + 1.17462 j)
(
)
= − 0.128259 m/s 2 k
Acceleration of point B.
a B = a B′ + a B/F + 2Ω × v B/F
(
) (
) (
)
a B = 0.381 m/s 2 i + 0.1069 m/s 2 j − 0.1283 m/s 2 k W
SOLUTION a P = a P′ + a P/F + ac
For each pin: Acceleration of the coinciding point P′ of the plate. For each pin, a P′ = rω 2 towards the center O. Acceleration of the pin relative to the plate.
For pins P1, P2 , and P4 ,
a P/F = 0
For pin P3 ,
a P/F =
u2 r
Coriolis acceleration ac . For each pin ac = 2ω u with ac in a direction obtained by rotating u through 90° in the sense of ω, i.e. . Then,
a1 = rω 2 a 2 = rω 2 a3 = rω 2 a 4 = rω 2
+ [ 2ω u + [ 2ω u u2 + r
+ [ 2ω u
]
a1 = rω 2i − 2ω uj W
]
a 2 = 2ω ui − rω 2 j W + [ 2ω u
]
]
u2 a3 = − rω 2 + + 2ω u i W r
(
)
a 4 = rω 2 + 2ω u j W
PROBLEM 15.188 A 60-mm-radius disk spins at the constant rate ω 2 = 4 rad/s about an axis held by a housing attached to a horizontal rod that rotates at the constant rate ω 1 = 5 rad/s. For the position shown, determine (a) the angular acceleration of the disk, (b) the acceleration of point P on the rim of the disk if θ = 0, (c) the acceleration of point P on the rim of the disk if θ = 90o.
SOLUTION ω = ω1i + ω 2k
Angular velocity.
ω = ( 5 rad/s ) i + ( 4 rad/s ) k (a) Angular acceleration. Frame Oxyz is rotating with angular velocity Ω = ω1i . =ω Oxyz + Ω × ω α =ω
= 0 + ω1i × (ω1i + ω 2k ) = −ω1ω 2 j = − ( 4 )( 5 ) j = −20 j
(
)
) (
)
(
)
α = − 20.0 rad/s 2 j W
(b) θ = 0. Acceleration at point P. rP = ( 60 mm ) i = ( 0.06 m ) i v P = ω × rP = ( 5i + 4k ) × 0.06i = 0.24 j a P = α × rP + ω × v P = −20 j × 0.06i + ( 5i + 4k ) × 0.24 j
= 1.2k + 1.2k − 0.96i = −0.96i + 2.4k
(
a P − 0.960 m/s 2 i + 2.40 m/s 2 k W
(c) θ = 90°. Acceleration at point P. rP = ( 0.06 m ) j v P = ω × rP = ( 5i + 4k ) × 0.06 j = −0.24i + 0.3k a P = α × rP + ω × v P = −20 j × 0.06 j + ( 5i + 4k ) × ( −0.24i + 0.3k ) = 0 + 0 − 1.5j − 0.96 j + 0 = −2.46 j a P = − 2.46 m/s 2 j W
PROBLEM 15.217 Y B
u
360 mm
The bent rod shown rotates at the constant rate ω 1 = 5 rad/s and collar C moves toward point B at a constant relative speed u = 975 mm/s. Knowing that collar C is halfway between points B and D at the instant shown, determine its velocity and acceleration.
C A 150 mm
E
520 mm
w1
X
D
7.
SOLUTION Geometry.
rB = (0.36 m)j,
rD = (0.15 m) i + (0.52 m)k
r B/D = – (0.15 m) i + (0.36 m) j – (0.52 m) k lDB =
>0.15C + >0.36C + >0.52C 2
2
2
= 0.65 m
rC = 1 (rB + rD) = (0.075 m) i + (0.18 m)j + (0.26 m) k 2 9 = w1k = (5 rad/s) k
Let frame Axyz rotate with angular velocity Velocity Analysis.
u = 0.975 m/s vC ¢ = 9 ´ rC = 5k ´ (0.075i + 0.18j + 0.26k) = – (0.9 m/s) i + (0.375 m/s)j
vC/F =
0.975 u rD/B = (– 0.15i + 0.36j – 0.52 k) = – (0.225 m/s)i + (0.54 m/s) j – (0.78 m/s) k 0.65 l DB vC = vC ¢ + vC/F vC = – (1.125 m/s)i + (0.915 m/s) j – (0.78 m/s)k J
Acceleration Analysis.
aC/F = 0
aC ¢ = 9 ´ vC ¢ = 5k ´ (0.9i + 0.375j) = – (1.875 m/s2) i – (4.5 m/s2) j aC/F = 0 29 ´ vC/F = (2) (5k) ´ (0.225i + 0.54j – 0.78k) = – (5.4 m/s2) i – (2.25 m/s2) j aC = aC ¢ + aC/F + 29 ´ vC/F a C = – (7.28 m/s2) i – (6.75 m/s2) j J
PROBLEM 15.237 In the position shown the thin rod moves at a constant speed u = 60 mm/s out of the tube BC. At the same time tube BC rotates at the constant rate ω 2 = 1.5 rad/s with respect to arm CD. Knowing that the entire assembly rotates about the X axis at the constant rate ω 1 = 1.2 rad/s, determine the velocity and acceleration of end A of the rod.
SOLUTION rA = (120 mm ) i + (180 mm ) k
Geometry. Method 1
Let the rigid body DCB be a rotating frame of reference. Its angular velocity is
Ω = ω1i + ω 2k = (1.2 rad/s ) i + (1.5 rad/s ) k.
Its angular acceleration is
α = ω1i × ω 2k = −ω1ω 2 j = − 1.8 rad/s 2 j.
(
)
Motion of the coinciding point A′ in the frame. v A′ = Ω × rA = (1.2i + 1.5k ) × (120i + 180k ) = −216 j + 180 j = − ( 36 mm/s ) j a A′ = α × rA + Ω × v A′ = ( −1.8j) × (120i + 180k ) + (1.2i + 1.5k ) × ( −36 j)
(
) (
)
= 216k − 324i − 43.2k + 54i = − 270 mm/s 2 i + 172.8 mm/s 2 k
Motion of point A relative to the frame. v A/F = ui = ( 60 mm/s ) i,
Velocity of point A.
a A/F = 0
v A = v A′ + v A/F v A = ( 60.0 mm/s ) i − ( 36.0 mm/s ) j W
Coriolis acceleration.
2Ω × v A/F
(
)
2Ω × v A/F = ( 2 )(1.2i + 1.5k ) × 60i = 180 mm/s 2 j
Acceleration of point A.
a A = a A′ + a A/F + 2Ω × v A/F
(
) (
) (
)
a A = − 270 mm/s 2 i + 180.0 mm/s 2 j + 172.8 mm/s 2 k W
PROBLEM 15.235 CONTINUED Method 2 Let frame Dxyz, which at instant shown coincides with DXYZ, rotate with an angular velocity Ω = ω1i = 1.2i rad/s. Then the motion relative to the frame consists of the rotation of body DCB about the z axis with angular velocity ω 2k = (1.5 rad/s ) k plus the sliding motion u = ui = ( 60 mm ) i of the rod AB relative to the body DCB.
Motion of the coinciding point A′ in the frame. v A′ = Ω × rA = 1.2i × (120i + 180k ) = − ( 216 mm/s ) j
(
)
a A′ = Ω × v A′ = 1.2i × ( −216 j) = − 259.2 mm/s 2 k
Motion of point A relative to the frame. v A/F = ω 2k × rA + ui = 1.5k × (120i + 180k ) + 60i = (180 mm/s ) j + ( 60 mm/s ) i a A/F = α 2k × rA + ω 2k × (ω 2k × rA ) + ui + 2ω 2 j × ( ui ) = 0 + 1.5k × 180 j + 0 + ( 2 )(1.5k ) × ( 60i )
(
) (
)
= 180 mm/s 2 j − 270 mm/s 2 i
Velocity of point A.
v A = v A′ + v A/F v A = −216 j + 180 j + 60i v A = ( 60 mm/s ) i − ( 36 mm/s ) j W
Coriolis acceleration.
2Ω × v A/F
(
)
2Ω × v A/F = ( 2 )(1.2i ) × (180 j + 60i ) = 432 mm/s 2 k
Acceleration of point A.
a A = a A′ + a A/F + 2Ω × v A/F a A = −259.2k + 180 j − 270i + 432k
(
) (
) (
)
a A = − 270 mm/s 2 i + 180.0 mm/s 2 j + 172.8 mm/s 2 k W
PROBLEM 15.233 The arm AB of length 5 m is used to provide an elevated platform for construction workers. In the position shown, arm AB is being raised at the constant rate dθ / dt = 0.25 rad/s; simultaneously, the unit is being rotated counterclockwise about the Y axis at the constant rate ω 1 = 0.15 rad/s. Knowing that θ = 20°, determine the velocity and acceleration of point B.
SOLUTION Frame of reference. Let moving frame Axyz rotate about the Y axis with angular velocity Ω = ω1j = ( 0.15 rad/s ) j.
rB/ A = −5cos 20°i + 5sin 20° j = − ( 4.6985 m ) i + (1.7101 m ) j
Geometry.
Place point O on Y axis at same level as point A.
rB/O = rB/ A + rA/O = rB/ A + ( 0.8 m ) i = − ( 3.8985 m ) i + (1.7101 m ) j
Motion of corresponding point B′ in the frame. v B′ = Ω × rB/O = ( 0.15 j) × ( −3.8985i + 1.7101j) = ( 0.58477 m/s ) k
(
)
a B′ = Ω × v B′ = ( 0.15j) × ( 0.58477k ) = 0.087715 m/s 2 i ω2 = −
Motion of point B relative to the frame.
dθ k = ( 0.25 rad/s ) k dt
v B/F = ω2 × rB/ A = ( −0.25 ) k × ( −4.6985i + 1.7101j) = − ( 0.42753 m/s ) i + (1.17462 m/s ) j
a B/F = ω2 × v B/F = ( −0.25k ) × ( −0.42753i + 1.17462 j)
(
) (
)
= 0.29365 m/s 2 i + 0.106881 m/s 2 j
Velocity of point B.
v B = v B′ + v B/F v B = − ( 0.428 m/s ) i + (1.175 m/s ) j + ( 0.585 m/s ) k W
Coriolis acceleration.
2Ω × v B/F = ( 2 )( −0.15j) × ( −0.42753i + 1.17462 j)
(
)
= − 0.128259 m/s 2 k
Acceleration of point B.
a B = a B′ + a B/F + 2Ω × v B/F
(
) (
) (
)
a B = 0.381 m/s 2 i + 0.1069 m/s 2 j − 0.1283 m/s 2 k W
