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Manual
Problem 180
Solutions 2.172
The reciprocating rectilinear motion mechanism shown consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between C and B . Letting D 30 , !AB D 50 rpm D constant, R D 0:3 ft, and h D 0:6 ft, determine P and R , i.e., the angular velocity and angular acceleration of the slotted arm CD , respectively.
Solution
Let R D AB and r D BC . The velocity of B in the .uR ; u / and .ur ; u / component OO OO systems are vB D R!AB u and vB D r ur C r P u : E O E PO O (1)
To nd P take advantage of the fact that for D 30 and R D 0:5h we have a right triangle ABC . Then ur D u and uR D u . Converting the u component of vB to O O O O O E uR and equating components we nd O r P uR D 0 O ) P D 0: (2)
The acceleration of B in the .uR ; u / and .ur ; u / component systems are OO OO aB D E
2 R!AB uR O
and
aB D r E R
r P 2 ur C r R C 2r P u : O PO
(3)
Convert the u component of aB to uR , equate components of Eq. (3), plug in Eq. (2), and substitute O E O p r D h2 R2 to nd rRD
2 R!AB
)
RDp
2 R!AB
h2
R2
D
15:8 rad=s2 .
200
Solutions Manual
Problem 2.190
An interesting application of the relative motion equations is the experimental determination of the speed at which rain falls. Say you perform an experiment in your car in which you park your car in the rain and measure the angle the falling rain makes on your side window. Let this angle be rest D 20 . Next, you drive forward at 25 mph and measure the new angle, motion D 70 , that the rain makes with the vertical. Determine the speed of the falling rain.
Solution
We will use a Cartesian Coordinate system where x and y represent the horizontal and vertical directions, respectively. The orientation of the angle allows us to write the velocity of the rain with respect to the stationary car as vR D E vR .sin rest { C cos rest | / : O O (1)
The second piece of information allows us to write the velocity of the rain with respect to the moving car as vR=C D E vR=C .sin motion { C cos motion | /: O O (2)
Relative kinematics tells us that we must have vR D vR=C C vC : E E E Substituting Eq. (1) and Eq. (2) into Eq. (3), vR sin rest { O vR cos rest | D vC { O O vR=C sin motion { O vR=C cos motion | : O (4) (3)
This vector equation is a system of two equations in two unknowns vR and vR=C vR sin rest D vC vR=C sin motion ; (5) (6)
vR cos rest D The solution to the system of equations yields vR D
vR=C cos motion ;
vC cos rest tan motion
sin rest
D 16:4 ft=s.
262
Solutions Manual
Problem 2.246
A jet is ying straight and level at a speed v0 D 1100 km=h when it turns to change its course by 90 as shown. The turn is performed by decreasing the paths radius of curvature uniformly as function of the position s along the path while keeping the normal acceleration constant and equal to 8g , where g is the acceleration due to gravity. At the end of the turn, the speed of the plane is vf D 800 km=h. Determine the radius of curvature f at the end of the turn and the time tf that the plane takes to complete its change in course.
Solution
The radii of curvature at the beginning and end of the turn are
2 v0 D 1190 m an 2 vf
0
D
and
f
D
an
D 629 m.
(1)
We must satisfy the condition .s/ D 0 C s , where is a constant to be determined. Manipulate the equation v D ! , where ! is angular velocity. See hint on next page v d d ds d d 1 for an explanation of !D D D Dv ) D: (2) dt ds dt ds ds what's going on here Integrate Eq. (2). Z
=2
0
d D
Z
sf
0
ds 0C s )
) D
2 2
D ln
1
ln
f 0
0
C sf
0
)
2
D
1
ln
f 0
:
D
0:4055: D
d ds
Knowing that
represents the rate of change of curvature with respect to position (
v2
) we can take the
derivative with respect to s of the equation an D v 2 D 8g ) 2v
and manipulate as follows: ) v dv D 4g : ds
dv d D 8g ds ds
However, v dv represents the quantity v . Therefore P ds dv D 4g dt ) Z
tf
0
dt D
Z
vf
v0
dv 4g
)
tf D
1 4g
vf
v0 D 5:24 s.
Hints for Problem 2.246, HW 2
This problem has a couple of challenges which involve switching back and forth between different descriptions of motion.
One important given piece of information is that and are at degrees. It is difficult to utilize this fact using path coordinates, but much easier in a well chosen Polar systems. The key is to choose a moving set of Polar coordinates whose origin starts at when the plane begins the turn, and ends at at the end of the turn. For family this of Polar coordinate systems, is always the same as the radius of curvature . Furthermore, the difference between the staring and final is exactly . What remains is to relate the various quantities between the Path systems and the Polar systems. For example, you know that
but to use that, you need to relate
to
and . The key is the following relation
By manipulating this, you can obtain a relationship between and . Note that you are also told that is uniformly decreasing as a function of , i.e. where is a constant that can be calculated from the given information.
where
is a constant that can be calculated from the given information.
The first hint above allows you to calculate as a function of . The problem requires you to determine the final time. Again, this is not immediate in Path coordinates, since quantities are normally expressed as a function of the path length rather than time. Therefore, to obtain time information, you need to figure out a relationship between and other known quantities such as , and . You are given the fact that normal acceleration is a constant , i.e.
Differentiating this relationship with respect to terms of quantities already calculated.
and observing that
will give
in
Page generated 2009-04-15 12:17:40 PDT, by jemdoc.
272
Solutions Manual
Problem 2.257
Block B is released from rest at the position shown and it has a constant accelerate downward aB D 5:7 ft=s2 . Determine the velocity and acceleration of block A at the instant that B touches the oor.
This should be m/s^2
Solution
The length of the rope is LD q
For t D 0 yA .0/ D l w . Let tf represent the time at the nal position. The length of the rope is expressed at the initial and nal positions as q q 2 2 L D d 2 C yA .0/ C yB .0/ and L D d 2 C yA .tf / C yB .tf /: Subtracting these two equations we get q q 2 2 C y 2 .t / d d 2 C yA .0/ C yB .tf / Af yB .0/ D 0:
2 d 2 C yA C yB :
(1)
The quantity yB .tf / yB .0/ is equivalent to h. Squaring both sides we obtain q q q 2 2 2 2 2 2 C y 2 .t / D 2 C y 2 .0/ d d h ) d C yA .tf / D d C yA .0/ 2h d 2 C yA .0/ C h2 : Af A r q 2 2 yA .tf / D yA .0/ C h2 2h d 2 C yA .0/ D 0:1864 m: Differentiate Eq. (1) and solve for yA . P yA yP A 0D q C yB P 2 d 2 C yA ) yA D P yB P yA q 2 d 2 C yA : (2)
Use the constant acceleration equation s 2 D s0 C 2ac .s P P2
The velocity of A when B hits the ground is
yB D 2aB h D 4:775 m=s: P2 yA .tf / D P 64:2 m=s:
s0 / to nd yB after B has traveled a distance h. P
This should be v_B = sqrt{2a_B h} = 4.775 m/s
Differentiate Eq. (2) with respect to time q q yB R yB yA PP 2 2 yA D R d 2 C yA C d 2 C yA 2 yA yA The acceleration of A when B hits the ground is yA .tf / D R 22;100 m=s2 :
q
2 d 2 C yA
yB yA PP
:
See an alternative and simpler solution method on next page
Dynamics
281
Problem 3.5
The motor M is at rest when someone ips a switch and it starts pulling in the rope. The acceleration of the rope is uniform and such that it takes 1 s to achieve a retraction rate of 4 ft=s. After 1 s the retraction rate becomes constant. Determine the tension in the cable during and after the initial 1 s interval. The cargo C weighs 130 lb, the weight of the ropes and pulleys is negligible, and friction in the pulleys is negligible.
Solution
Balance Principles: X Fy W 3T C W D W aC : g (1)
Kinematics
Differentiate Eq. (2) with respect to time to obtain relations between vC and the retraction P R rate of the rope L, and aC and the change in retraction rate L as P L D 3vC ; and R L D 3aC : (3)
L D 3yC C constants:
(2)
R The motor accelerates uniformly from rest, which implies that L is constant. P P R , where the initial rate L.0/ D 0 ft=s. Since L.t D P P Then L.t/ D L.0/ C Lt 1 s/ D 4 ft=s (which is negative because the rope is being retracted), we have R L D 4 ft=s2 . Combining Eqs. (1) and (3), the tension of the rope in terms of R L is ! R W L TD 1 D 45:1 lb: (4) 3 3g R To nd the tension in the cable after the motor is nished accelerating we set L equal to 0 in Eq. (4): TD W D 43:3 lb: 3
326
Solutions Manual
Problem 3.50
Derive the equations of motion for the pendulum supported by a linear spring of constant k and unstretched length ru . Neglect friction at the pivot O , the mass of the spring, and air resistance. Treat the pendulum bob as a particle of mass m and use polar coordinates.
Solution
Based on the FBD shown at the right, the Newton-Euler equations for this system are X Fr W mg cos Fs D mar ; (1) X F W mg sin D ma ; (2) ru /. The kinematic equations are given by P r 2; and R a D r C 2r : PP ar D r R (3)
We have the material equation Fs D k.r
Substituting Eqs. (3) into Eq. (1) and Eq. (2) gives the desired equations of motion as r R P r2 C k .r ru / g cos D 0; m R r C 2r C g sin D 0: PP (4) (5)

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