INTRODUCTION
1.9 FURTHER EXAMPLES
(d) (c)
v (b)
(a)
v
v
FIGURE 1.22
(a) A suspension system for a small vehicle such as a golf cart is the second benchmark problem. (b) In early chap- ters, the golf cart is modeled as a SDOF system. (c) The analysis grows in complexity as the chapters progress. In later chapters, the mass of the wheel is taken into account. (d) The distribution of mass on the body is considered.
E X A M P L E 1 . 8
The slender bar of Example 1.4 and Figure 1.16 is pinned at Aand held in the horizon- tal position by a cable. The cable is cut at t⫽0.
(a) What is the bar’s angular velocity after it has rotated through 10°?
(b) What are the reactions at the pin support after it has rotated through 10°?
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S O L U T I O N
(a) Let position 1 refer to the bar immediately after the cable is cut. Let position 2 refer to the bar after it has rotated through 10°. All external forces are conservative; thus, conserva- tion of energy applies between positions 1 and 2 as
(a) Take the datum for potential energy calculations for the gravity force to be position 1, then V1⫽0, and The kinetic energy in position 1 is zero, and
(b) Kinematics (the relative velocity equation) is used to relate the velocity of the mass center to the angular velocity of the bar so that Substituting into Equation (a), we have
(c) which is solved to yield
(d) (b) Summing moments about the pin support on the free-body diagrams after the body has rotated through 10° are illustrated in Figure 1.23. Taking moments about the pin support yields which is the same as the initial value. This is to be expected, as the external forces are constant, which implies uniformly accelerated motion. Summing forces using the free-body diagrams according to give
(e) By equating coefficients of the unit vectors, the reactions are determined as
(f) Ry = mg a1 - (g)
4
7cos10° + 8
7sin210°b = 0.472mg Rx = -
4mg
7 sin10°(1 + 2cos10°) = -0.295mg + mL
3a24g
7L sin10°b(-cos10°i + sin10°j) Rxi + (Ry - mg)j = mL
3a12g
7Lb(-sin10°i - cos10°j) (gF)ext = (gF)eff a = 12g
7L, v =
A 24g
7L sin10° = 0.818A g L 0 = 1
2maL
3vb2 + 1 2a 1
12mL2bv22 - mgL
3 sin10°
v = L
3v.
T2 = 1
2mv22 + 1 2a 1
12mL2bv22 V2 = - mgL
3 sin10°.
T1 + V1 = T2 + V2
mg Ry Rx
External forces
m α
Effective forces L 3
mL2α 1 12 mLω2
3
FIGURE 1.23
FBDs after bar of Example 1.8 has rotated through10°.
Determine the acceleration of the block Figure 1.24(a).
S O L U T I O N
The acceleration of the block is assumed to be upward, which is consistent with the assumed direction of the angular acceleration of the disk. The point on the disk where the cable is in contact with it has the same acceleration (r) as the cable. Assuming the cable is inex- tensible, it has the same acceleration as the block. Summing moments about the mass center by applying to the FBDs shown in Figure 1.24(b) leads to
(a) Solving for gives
(b) The acceleration of the block is
(c) a = ra = (0.3 m)(68.5 rad/s2) = 20.5 m/s2
a =
M - mgr I + mr2 =
(18 N
#
m) - (1.3 kg)(9.81 m/s2)(0.3 m)0.09 kg
#
m2 + (1.3 kg)(0.3 m)2 = 68.5 rad/s2 M - mgr = mra(r) + Ia(gMO)ext = (gMO)eff
E X A M P L E 1 . 9
0 r
m M
(a)
M = 18 N · m m = 1.3 kg r = 30 cm I = 0.09 kg · m2
R
mg M
(b)
External forces Effective forces
=
mpg
mrα Iα
FIGURE 1.24
(a) System of Example 1.9. (b) FBDs drawn at an arbitrary instant showing the external forces and the effective forces.
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A thin disk of mass 5 kg, radius 20 cm, and attached to a spring of stiffness 2000 N/m is in equilibrium when it is subject to an applied force P⫽10 N. The coefficient of friction between the disk and the surface is 0.1.
(a) What is the maximum displacement of the disk from its equilibrium position, assum- ing no slipping between the disk and the surface?
(b) What is the angular acceleration of the disk immediately after it reaches its maximum displacement?
(c) Is the no-slip assumption correct?
S O L U T I O N
(a) Let position 1 refer to the position when the disk is in equilibrium, and let position 2 refer to the position when the disk reaches its maximum displacement. Application of the principle of work and energy between position 1 and position 2 for the disk gives
(a) The kinetic energy of the disk in position 1 is zero, because the disk is at rest. The kinetic energy of the disk in position 2 is zero, because the disk reaches its maximum displacement.
The only source of potential energy is the spring force. The potential energy in the spring in position 1 is zero, as the spring is unstretched. Letting xbe the maximum displacement, the potential energy in position 2 is
(b) The friction force does no work, since the disk rolls without slipping. Thus, the velocity of the point where the friction force is applied is zero. The only non-conservative force is the applied force P. Its work is
(c) Substituting into Equation (a),
(d) or
(e) (b) Summing moments about the contact point as and using the free-body diagrams drawn immediately after the disk reaches its maximum displacement (illustrated in Figure 1.25) yields
(f) If the disk rolls without slipping, the velocity of the point of contact is identically zero, and its acceleration only has an upward component of r2. Application of the horizontal
-kxr + Pr = 1
2mr2a + mar
(gMO )ext = (gMO )eff x =
2P k =
2(10 N)
2000 N/m = 0.01 m Px = 1
2kx2 U1:2NC =
L
x 0
Pdx = Px V2 = 1
2kx2
T1 + V1 + U1:2NC = T2 + V2
E X A M P L E 1 . 1 0
component of the relative acceleration equation between the point of contact and the mass center yields . Substituting this result into Equation (b) leads to
(g) (c) Summing moments about the mass center as and using the free-body diagrams of Figure 1.25 yields
(h) The maximum value of from when the motion is initiated to when the disk reaches its maximum displacement should be used in the calculation. The maximum value occurs in position 1 when
(i) and
(j) The maximum available friction force is mg ⫽0.1(5 kg) (9.81 m/s2) ⫽ 4.91 N. Since the friction force is less than the maximum allowable friction force, the disk rolls without slipping.
F = 1
2mra = 1
2(5kg)(0.2 m)(6.67 rad/s2) = 3.33N a =
2P 3mr =
2(10N)
3(5 kg)(0.2 m) = 6.67 rad/s2 Fr =
1
2mr2aQF = 1 2mra
(gMC)ext = (gMC)eff a =
2(P - kx)
3mr =
2310 N - (2000 N/m)(0.01 m)4
3(5 kg)(0.2 m) = 6.67 rad/s2 a = ra
N
F
P P
External forces Effective forces kx
mr2α
ma = mrα
mg 1
2
=
FIGURE 1.25
FBDs of system in Example 1.10.
Summing moments about the point of contact helps to solve for the angular acceleration assuming no slipping. Summing moments about the mass center finds the friction force which is checked against the maximum value to determine if slipping occurs.
E X A M P L E 1 . 1 1
A baseball player holds a bat with a centroidal moment of inertia a distance afrom the bats mass center. His “bat speed” is the angular velocity with which he swings the bat. The pitched ball is a fastball which reaches the batter with a velocity v. Assuming his swing is a rigid-body rotation about an axis perpendicular to his hands, where should the batter hit the ball to minimize the impulse felt by his hands?
S O L U T I O N
When the better hits the ball, it exerts an impulse on the bat: call it B. Since the batter is holding the bat, he feels an impulse as he hits the ball: call it P. The effect of hitting the
I
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ball is to change the bat speed from 1to2. The impulse momentum diagrams of the bat during the time are shown in Figure 1.26.
Applying the principle of linear impulse and momentum to Figure 1.26 leads to (a) Application of the principle of angular impulse and angular momentum about an axis through the batter’s hands yields
(b) Solving Equation (b) for B, we have
(c) Substituting Equation (c) into Equation (a) and solving for Pleads to
(d) Thus, P⫽0 if
(e) Thus, the angular impulse felt by the batter is zero if bsatisfies Equation (e). The location of bis called the center of percussion.
b = a + I ma
P = (v1 - v2)aI + ma2 b - mab B =
(I + ma2)
b (v2 - v1)
Iv1 + mav1(a) - B(b) = Iv2 + mav2(a) mav1 + P - B = mav2
b
B
= +
= +
a
P
Momenta of bat immediately before striking ball
External impulses during striking ball
System momenta immediately after striking ball
Iω1 Iω2
maω2
maω1
FIGURE 1.26
Impulse momentum diagrams for Example 1.11 as batter hits ball.