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http://www.book4me.xyz/solution-ma http://www.boo k4me.xyz/solution-manual-mechanical-vibr nual-mechanical-vibration-palm/ ation-palm/ ˙ , and the tangential tangential velocity component of the cylinder is R is R φ velocity component of the contact point is r is r θ˙. If there is no slipping, these two components ˙ = r must be equal. Thus R Thus R φ = r θ˙. Problem Problem 2.1 The
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˙ , and the tangential tangential velocity component of the cylinder is R is R φ velocity component of the contact point is r is r θ˙. If there is no slipping, these two components ˙ = r must be equal. Thus R Thus R φ = r θ˙. Problem Problem 2.2 The
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Problem 2.3 Let
θ be the angle of the line B line BA A measured from the vertical. That is, θ is, θ = = 0 ˙ ¨ when point A point A is in contact with the surface. Then ω = θ and α = θ. Let s be the linear displacement of point B point B .. Then s = Rθ = Rθ vB = R θ˙ = Rω = Rω aB = R θ¨ = Rα = Rα Establish an xy coordinate coordinate system whose origin is located at the point of contact of point A with the surface at time t = 0. The coordinate x is positive to the left and y is positive upward. Then the xy coordinates of point A point A are x = s = s
= R((θ − sin θ) − R sin θ = R x˙ = R = R θ˙(1 − cos θ cos θ)) = v B (1 − cos θ cos θ))
y = R = R
− R cos θ
y ˙ = R = R θ˙ sin θ sin θ = v = v B sin θ sin θ
So the velocity of point A has the compone components nts x˙ and y ˙ . The magnitude of the velocity is vA =
2
2
x˙ + y ˙ =
2
vB (1
2
cos θ)) − cos θ
2
2
+ vB sin θ = v = v B 2(1
To find the acceleration, differentiate again to obtain x¨ = ˙Bv(1
cos θ)) + v + vB ˙θ sin θ sin θ = a = a B (1 − cos θ cos θ)) + Rω − cos θ
2
cos θ)) − cos θ
sin θ
y y¨ = = ˙Bv sin θ sin θ + + vB ˙θ cos θ cos θ = a = a B sin θ sin θ + + Rω Rω 2 cos θ or x¨ = Rα = Rα(1 (1
+ Rω − cos θ) + Rω
2
sin θ sin θ
y y¨ = = Rα sin θ sin θ + + Rω Rω 2 cos θ The acceleration magnitude is aA =
x ¨2 + y y¨ 2 =
2R2α2 + R2 ω 4 + 2R 2 R2αω 2
2
2
− 2R α
cos θ cos θ
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Problem 2.4 Let x Let x be be
the horizontal displacement of the vehicle measured to the left from directly directly underneath underneath the pulley. pulley. Let y be be the height height of the block above the ground. ground. Let D be the length of the hypotenuse of the triangle whose sides are x and 3 m. Then, from the Pythagorean theorem, D2 = x 2 + 32 (1) Differentiate this to obtain
DD˙ = x = x x˙
The total cable length is 10 m, so D + D + 3
(2)
= 10 or − y =
D = y = y + 7
(3)
This gives x2 = D 2 and
2
+ 7) − 9 − 9 = (y +
˙ = xx˙ = xx˙ y ˙ = = D D 7 + y
(4) (5)
Substituting the given values of y = 2, x˙ = 0.2 into Equations (4) and (5), we obtain x = 72 = 6 2 and 6 2(0. 2(0.2) y ˙ = = = 0.1885 1885 m/s 9 This is the velocity of the block after it has been raised 2 m. To obtain the acceleration, differentiate Equations (2) and (5) to obtain
√
√
√
¨ = ˙ 2x+ x˙ x D˙ 2 + D˙ D ¨ ¨ y y¨ = =D ¨: Solve for D 2 ¨ ¨ = x˙ + x˙ x D D˙
− D˙
2
=
(0. (0.2)2 + 0 (0. (0.1885)2 = 0.0237 0237 m/s2 0.1885
−
¨ , the acceleration of the block after it has been raised 2 m is 0.0237 m/s 2 . Since y y¨ = =D
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Problem Problem 2.5 Only
one of the pulleys translates, so v so v 2 = v = v 1 /2.
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Let vC denote denote the veloci velocity ty of the middle middle pulley pulley. Then Then vC = vA /2 and vB = v C /2. Thus, v Thus, v B = (vA /2)/ 2)/2 = v = v A /4.
Problem Problem 2.6
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Problem Problem 2.7
M O = I O α = I = I O θ¨ 2 I O = I RG + mRL2 + mC LC RG + m
M O = Thus
gL sin θ − mC gL C sin θ −mRgL sin
2 ¨ (I RG + mR L2 + mC LC )θ = RG + m
sin θ −(mRL + mC LC )g sin θ
If m m R
I RG ≈ 0 and if I RG ≈ 0, then 2 ¨ mC LC θ =
or
−mC LC g sin θ
LC θ¨ =
−g sin θ
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˙ , and the tangential tangential velocity component of the cylinder is R is R φ velocity component of the contact point is r is r θ˙. If there is no slipping, these two components ˙ = r must be equal. Thus R Thus R φ = r θ˙ and Rφ¨ = r = r θ¨. Problem Problem 2.8 The
1 ˙ 2 1 KE = I G φ + M (r 2 2
P E = = M g [(r [(r
− R)θ˙
2
)cos θ]] = M g(r − R) (1 − cos θ) − R) − (r − R)cos θ
Because K E + + P E = = constant, then d(KE + + P E )/dt = /dt = 0 and ˙ φ¨ + M ( I G φ M (r
+ M g (r − R)sin θ )sin θ θ˙ = 0 − R) ˙θθ¨ + M 2
˙ = r ¨ . Thus But R But R φ = r θ˙ and φ¨ = r = r θ/R. θ/R I G
r R
2
θ˙θ¨ + M + M ((r
− R) ˙θθ¨ + M + M g(r − R)sin θ )sin θ θ˙ = 0 2
Cancel θ˙ and collect terms to obtain
I G
r R
2
+ M (r
2
− R)
¨ θ + M g (r
)sin θ = 0 − R)sin θ
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˙ , and the tangential tangential velocity component of the cylinder is R is R φ velocity component of the contact point is r is r θ˙. If there is no slipping, these two components ˙ = r must be equal. Thus R Thus R φ = r θ˙ and Rφ¨ = r = r θ¨. The energy method is easier than the force- moment method here because 1) the motions of the link and cylinder are directly coupled (i.e. if we know the position and velocity of one, we know the position and velocity of the other), 2) the only external force is conservative (gravity), and 3) we need not compute the reaction forces between the link and the cylinder. Problem Problem 2.9 The
1 1 K E = = I G ˙φ2 + M (r 2 2
P E = = M g(r
−
R)θ˙
2
1 L˙ + m θ 2 2
2
1 L + I O + m + m 2 2
2
θ˙2
L cos θ)) + mg( mg (r − )(1 − cos θ) − R) (1 − cos θ 2
Because K E + + P E = = constant, then d(KE + + P E )/dt = /dt = 0, and
L2 ˙ ¨ L 2 ˙ ¨ ˙ ¨ I G φφ+M (r R) θ θ+m θ θ+ I O + m 4 2
−
2
L sin θ θ˙ = 0 2
−
θ˙θ¨+M g (r R)sin θ )sin θ θ˙+mg r
−
˙ ¨ . Substitute But φ˙ = r θ/R and φ¨ = r θ/R. θ/R Substitute these expressions, expressions, cancel θ˙, and collect terms to obtain
I G
r R
2
+ M ( M (r
−
L2 L2 ¨ 2 R) + m + I O + m + m θ + M g (r 4 4
− R) + mg
L 2
− r
sin θ sin θ = 0
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Problem Problem 2.10 a)
Let point O be the pivot point and G be the center of mass. Let L be the distance from O to G to G.. Treat the pendulum as being composed of three masses: 1) m 1) m 1 , the rod mass above point O point O,, whose center of mass is 0.03 m above point O point O;; 2) m 2) m 2 , the rod mass below point O point O,, whose center of mass is 0.045 m below point O, O , and 3) m 3) m 3 , the mass of the 4.5 kg block. Then, summing moments about G gives m1 g (L + 0. 0 .03)
(0.045 − L) − m g (0. (0.09+ 0. 0.015 − L) = 0 − m g(0.
where
2
3
0.06 1.4 = 0.56 kg 0.15 0.09 m2 = 1.4 = 0.84 kg 0.15 m3 = 4.5 kg m1 =
The factor g factor g cancels out of the equation to give 0.56(L 56(L + 0. 0.03)
84(0.045 − L) − 4.5(0. 5(0.105 − L) = 0 − 0.84(0.
which gives L gives L = = 0.084 m. b) Summing moments about the pivot point O point O gives I O θ¨ =
mgL sin θ −mgL sin
where m is the total mass. From the parallel-axis parallel-axis theorem, theorem, treating treating the rod as a slender slender rod, we obtain I O =
1 (1. (1.4)(0. 4)(0.06 + 0. 0.09)2 + (1. (1.4)(0. 4)(0.015)2 + (4. (4.5)(0. 5)(0.09 + 0.015)2 = 0.0525 0525 kg m2 12
·
and mg and mg L = (1. (1.4 + 4. 4.5)(9. 5)(9.81)(0. 81)(0.084) 084) = 4. 4.862 N m. Thus the equation of motion is
·
0.0525θ¨ = or
sin θ −4.862 sin θ
θ¨ + 92. 92 .61 sin θ = 0
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Problem 2.11 See
the figure for the coordinate definitions and the definition of the reaction force R. Let P be the point on on the axle. axle. Note Note that that y P coordinat nates es of the mass P = 0. The coordi center of the rod are x G = x P (L/2) L/2) sin θ and y and y G = (L/2) L/2) cos θ cos θ.. Thus
−
−
− L2 θ¨ cos θ + L2 θ˙
2
x ¨G = x ¨P
sin θ sin θ
L¨ L θ sin θ sin θ + + θ˙2 cos θ 2 2 Let m be the mass of the rod. Summing forces in the x direction: y y¨G =
mx ¨G = f = f
or
m x ¨P
−
L¨ L θ cos θ + θ˙2 sin θ sin θ = f = f 2 2
(1)
Summing forces in the y direction: y direction: my y¨G = R
− mg
or
L¨ L m θ sin θ sin θ + + θ˙2 cos θ = R = R 2 2
− mg
(2)
Summing moments about the mass center: I G θ¨ = (f L/2) L/2) cos θ cos θ (RL/2) RL/2) sin θ sin θ.. Substituting 2 for R from (2) and using the fact that I that I G = mL /12, we obtain
−
1 fL cos θ cos θ mL2 θ¨ = 12 2
−
mgL sin θ 2
−
mL2 sin2 θ ¨ θ 4
−
mL2 ˙2 sin θ cos θ θ sin θ 4
(3)
The model consists of (1) and (3) with m with m = = 20 kg and L and L = = 1.4 m.
Figure : for Problem 2.11
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Problem Problem 2.12
a) 2kx = kx = mg mg,, so x = mg = mg/ /2k.
b) T 0 = 0 1 mv 2 2 From conservation of energy, T 1 =
V 1 =
V 0 =
1 k(0. (0.06)2 2
1 k[0. [0.06 + 2(0 2(0..09)]2 2
mg (0..09) − mg(0
T 0 + V 0 = T = T 1 + V 1 Thus
1 2 1 1 mv = T 0 + V + V 0 V 1 = k(0. (0.06)2 k[0. [0.06 + 2(0 2(0..09)]2 + mg(0 mg (0..09) 2 2 2 Solve for v for v using m = 30 and k = 600 to obtain v obtain v = 0.828 m/s.
−
−
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Problem Problem 2.13 a)
From conservation of energy, T 0 + V + V 0
Thus
− W
01 01
= T = T 1 + V 1
0 + mgd sin mgd sin θ
1 − µmgd cos µmgd cos θ = mv 2
− µ cos θ) = 2.04
Thus v =
2gd(sin gd (sin θ θ
2
+0
m/s
b) From conservation of energy, T 1 + V + V 1 Thus
− W
12 12
1 mv12 + mgx + mgx sin θ 2
= T = T 2 + V 2
1 cos θ = 0 + kx − µmgx cos θ 2
2
This gives 6.2421 2421 + 20. 20.807x 807x which has the roots x roots x = 0.1249 and x = the spring is compressed by 0. 0 .1249 m.
2
3228x = 500x 500x − 8.3228x
−0.0999. Choosing the positive root, we see that
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Problem Problem 2.14
W = f x or W = W = 400(2)
−
−
1 k (x + x0)2 2
1 44 (2. (2.5)2 2
− (0. (0.5)
−x
−
2
2 0
−
mgx sin mgx sin θ
11(9. 11(9.81)2(0. 81)2(0.5) = 560. 560.09 N m
·
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Problem Problem 2.15
Let x Let x 1 be the initial stretch in the spring from its free length. Then (L + x1 )2 = (D1
2
− r)
+ D22
where r where r is the radius of the cylinder. This gives x gives x 1 = 1.828 m. From conservation of energy, T 1 + V 1 = T = T 2 + V 2 or
1 1 1 1 0 + kx21 = mv 2 + I ω 2 + kx22 2 2 2 2 where v where v = Rω = Rω and x and x 2 = D = D 1 r L = 1 m. This This gives gives
− −
0 + 41 41.77 =
1 1 10(0. 10(0.25ω 25ω )2 + 0.4ω 2 + 12. 12 .5 2 2
or ω = 7.56 rad/s.
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Problem Problem 2.16 The
velocity of the mass is zero initially and also when the maximum compression is attained. Therefore ∆T ∆T = = 0 and we have ∆T ∆T + + ∆V = ∆T + + ∆V s + ∆V g = 0 or ∆V s + ∆V g = 0. That That is, if the mass mass is drop dropped ped from from a heig height ht h above the middle spring and if we choose the gravitational potential energy to be zero at that height, then the maximum spring compression x compression x can can be found by adding the change in the gravitational potential energy 0 W ( W (h + x + x)) = W ( W (h + x + x)) to the change in potential energy stored in the springs. Thus, letting W letting W = mg, mg ,
−
−
1 k1(x2 2
W (h + x)] − 0) + [0[0 − W (
if x < d
where d where d is is the difference in the spring lengths (d ( d = 0.1 m). This gives gives the following following quadratic quadratic equation to solve for x: x : 1 k1x2 2 If x x
− Wx − Wh = 0
if x < d
(1)
≥ d, ∆V s + ∆V g = 0 gives 1 k1 (x2 2
(2k ) − 0) + 21 (2k 2
(x
2
W (h + x) x)] = 0 − d) − 0 + [0 − W (
if x
≥d
which gives the following quadratic equation to solve for x. x . (k1 + 2k 2 k2)x2
2
(2W + 4k 4k d)x + 2k 2 k d − 2W h = 0 − (2W 2
2
if x
≥d
(2)
For the given values, equation (1) becomes 104x2
200(9.81)x 81)x − 200(9. 200(9.81)(0. 81)(0.75) = 0 − 200(9.
if x < 0. 0 .1
which has the roots x = 0.494, which is greater than 0. 0.1, and x = 0.2978, which is not feasible. feasible. Thus, Thus, since there is no solution solution for which which x < 0.1, the side springs will also be compressed. From equation (2)
−
2.6
4
2
3200)x x + 160 − 1471. 1471.5 = 0 × 10 x − (1962 + 3200) which has the solutions: x = 0.344 and x and x = = −0.146. We discard the second solution because it is negati negative ve.. So the outer outer springs springs will be compre compresse ssed d by 0. 0.344 − 0.1 = 0.244 m and the middle spring will be compressed 0. 0.344 m.
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Problem Problem 2.17 From
conservation of angular momentum, H O1 = H = H O2
or
mv1 (0. (0.5) = I G + m + m2(1)2 + m1(0. (0.5)2 ω or
1 0.8(10)v 8(10)v1(0. (0.5) = (4)(1)2 + 4 + 0. 0.8(0. 8(0.5)2 ω 12
Solve for ω for ω:: ω = 0.882 882 rad rad/s
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Problem Problem 2.18 From
the figure, vB 1 = 10 cos 30 vA2 vB 2 0.6ω vB 2 e = 1 = = vB 1 vA1 vB 1 0 ◦
− −
− −
vB 1 = 10 cos 300 = 0.6ω
− vB
Thus
2
and vB 2 = 0.6ω
− 8.66
From conservation of momentum, H O1 = H = H O2 or m1 vB 1(0. (0.6) = I = I O ω + m + m1 vB 2(0. (0.6) where I O = I G + m + m2(0. (0.6)2 =
1 m2 (0. (0.8)2 + 4(0. 4(0.36) = 1. 1.653 12
Thus 0.8(8. 8(8.66)(0. 66)(0.6) = 1. 1.653ω 653ω + 0. 0 .8(0. 8(0.6)(0. 6)(0.6ω
− 8.66)
This gives ω = 4.28 rad rad/s
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Problem Problem 2.19 From
conservation of angular momentum, H O1 = H = H O2
or
mv1 (2L (2L) = I G + m2 L2 + m1 (2L (2L)2 ω where I G = Thus
1 1 m2 (2L (2L)2 = 4.5(2. 5(2.4)2 = 2. 2 .16 12 12
2
2
0.005(365)(2. 005(365)(2.4) = 2.16 + 4. 4.5(1. 5(1.2) + 0. 0.005(2. 005(2.4) ω Solve for ω for ω::
ω = 0.505 505 rad rad/s
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Problem 2.20 Let f n be
the reaction force on the pendulum from the pivot in the normal direction. When θ When θ = π = π/ /2, f 2, f n is vertical and is positive upward. Let f Let f t be the reaction force on the pendulum from the pivot in the tangential direction. When θ When θ = π = π/ /2, f 2, f t is horizontal and is positive to the right. Summing moments about the pivot at point O gives the equation of motion of the pendulum. L I O θ¨ = mg = mg cos θ cos θ 2 Summing moments about the mass center gives I G θ¨ = f = f t
L 2
Comparing these two expressions, we find that f t =
mgI G cos θ cos θ I O
Thus the tangential reaction force is f t = 0 when θ = π = π/ /2. To compute the normal reaction force, sum forces in the vertical direction to obtain man = f n
− mg
where the normal acceleration is given by the following expression for circular motion. an =
L 2 ω 2
Thus
L f n = mg = mg + + ma man = mg + mg + m m ω 2 2 We can compute ω either from energy conservation or by integrating the equation of motion. With the energy method we use the fact that the kinetic energy at θ = 90 equals the original potential energy at θ at θ = 0. Thus ◦
1 L I O ω 2 = mg 2 2 Since I O = mL = mL 2 /3 for a slender rod, this gives ω gives ω 2 = 3g/L. g/L . (Continued on the next page)
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