Upload
clinthopkins
View
219
Download
0
Embed Size (px)
Citation preview
8/12/2019 MT 2 Review
1/20
EGR 2513.001 and 002
MT Exam #2Spring 2014
Review Problems
8/12/2019 MT 2 Review
2/20
Review Problems
Given: At the instant shown, boat Atravels with a speed of 15m/s, which is decreasing at 3
m/s2, while boat B travels
with a speed of 10 m/s,which is increasing at 2 m/s 2.
Find:
The velocity and accelerationof boat A with respect to boatB at this instant.
Example 1 (Section 16.8)20 m
8/12/2019 MT 2 Review
3/20
Example 1 (continued)
Solution:Reference Frame: The xyzrotating reference frame isattached to boat B and
coincides with the XYZ fixedreference frame at the instantconsidered, Fig. a . Sinceboats A and B move along the
circular paths, their normalcomponents of acceleration are = = 1
0= 4.5 /
and =
= 10
0= 2.0 / .
Review Problems
8/12/2019 MT 2 Review
4/20
Thus, the motion of boats A and B with respect to the XYZframe are
= 15 / = 10 / = 4.5 3 = 2 2 /
Example 1 (continued)
Solution:
Also, the angular velocity and angular acceleration of thexyz reference frame with respect to the XYZ referenceframe are
=
= 10
0= 0.2 / = 0.2 /
= = 0 = 0.04 / = 0.04 /
Review Problems
8/12/2019 MT 2 Review
5/20
Example 1 (continued)
Solution:
And the position of boat A with respectto B is / = 20
Velocity: Applying the relative velocity equation,
= + / + 15 = 10 + 0.2 20 + 15 = 14 +
= 29 / Acceleration: Applying the relative acceleration equation,
= + / + 2 + 4.5 3
= 2 2 + 0.04 20 + 0.2 0.2 20+ 2 0.2 29 + = 8.8 2.8 +
= 4.3 0.2 /
Review Problems
8/12/2019 MT 2 Review
6/20
Given:The arm is rotating at a rate of = 5 rad/s when = 2 rad/s 2 and = 90 .
Find:The normal force it must exerton the 0.5-kg particle if theparticle is confined to move
along the slotted path definedby the horizontal hyperbolicspiral = 0.2 m.
Example 2 (Section 13.6)
Review Problems
8/12/2019 MT 2 Review
7/20
Example 2 (continued)
Solution:
= rad
= 5 rad/s
= 2 rad/s 2 = 0.2 = 0.12732 m = 0.2 = 0.40528 m/s = 0.2 2 3
+ = 2.41801 m/s 2
=
= 2.41801 0.12732 5 = 0.7651 m/s 2
= + 2 = 0.12732 2 + 2 0.40528 5 = 3.7982 m/s 2
Review Problems
8/12/2019 MT 2 Review
8/20
Example 2 (continued)
tan = =0.2 0.2
=
= tan 1
2= 57.5184
+ = : cos 32.4816 = 0.5 0.7651 = : + sin 32.4816 = 0.5 3.7982 +
Hence,= 0.453 N
= 1.66 N
= -57.518432.4816
F
N P
F
N P
Therefore, both N P and F are actingopposite to the assumed directions.
Why?
r
Review Problems
8/12/2019 MT 2 Review
9/20
Example 3 (Section 13.6)
Given:The pilot of an airplaneexecutes a vertical loop whichin part follows the path of a
cardioid, = 600 1 + cos ft.His speed at A = 0 is aconstant = 80 ft/s. Heweighs 150 lb f .
Find:The vertical force the seat belt must exert on him to holdhim to his seat when the plane is upside down at A.
Review Problems
8/12/2019 MT 2 Review
10/20
Example 3 (continued)
Solution:r = 600 1 + cos | =0 = 1200 ft = 600 sin | =0 = 0 = 600 sin 600 cos | =0 = 600
= +
(1)
80 = 0 + 1200
yields
= 0.06667
Also by taking the time derivative of Eq. (1)2 = 2 + 2 +
0 = 0 + 0 + 2 yields
= 0
Review Problems
8/12/2019 MT 2 Review
11/20
Example 3 (continued)
Thus,= = 600 0.06667 1200 0.06667 = 8 ft/s 2 = + 2 = 0 + 0 = 0
Therefore,
+ = : 150 = 1 03 .
8yields
= 113 lb f
Review Problems
8/12/2019 MT 2 Review
12/20
Example 4 (Section 17.5)
Given:The spool has a mass of100 kg and a radius ofgyration = 0.3 m. Thecoefficients of static andkinetic friction at A are and
= 0.2 and = 0.15 .
Find:The angular acceleration of the spool if P = 600 N.
Review Problems
8/12/2019 MT 2 Review
13/20
8/12/2019 MT 2 Review
14/20
Example 4 (continued)
Now, lets consider the effect of two potential errors in settingup this problem.
Case 1 Friction Force in Wrong DirectionConsider the following FBD:
F A
Still assuming = 0.4 but that F A is in theopposite direction, Eqs. (1) and (3) become,
600 = 40 (1a)150 + 0.4 = 9 (3a)
Solving these simultaneously, we have
= 15.6 rad/s 2 = 6.24 m/s 2 = 24.0 N Which is identical to our original answer since it shows thatF A was chosen in the wrong direction. Therefore, this error
is self-correcting.
Review Problems
8/12/2019 MT 2 Review
15/20
Example 4 (continued)Case 2 and in Inconsistent Directions
Assuming = 0.4 (i.e., and are in inconsistentdirections), Eqs. (1) and (3) become,
600 + = 100 = 40 (1b)150 0.4 = 9 (3b)
Solving these equations simultaneously, we have= 12.2 rad/s 2 = 4.88 m/s 2 = 600.0 N
Since > , the no-slip assumption is incorrect and,therefore, = = 0.15 = (0.15)(981) = 147.15 N.Substituting this result into Eqs. (1 b) and (3b), we have
= 18.68 rad/s (1c)
= 10.13 rad/s (3c)
Review Problems
8/12/2019 MT 2 Review
16/20
Example 4 (continued)Case 2 (continued)
Therefore, there is no solution (i.e. the two lines do notintersect) and, not surprisingly, the final results do not agreewith the correct solution. Therefore, the moral to this case isto follow the admonition given in the Section 17.5 Lecture,namely,
Be consistent in using the assumed directions. Thedirection of a G must be consistent with .
Review Problems
8/12/2019 MT 2 Review
17/20
Example 5 (Section 17.5)
Given: Assume the 50-lb roller is auniform cylinder and thefriction coefficients betweenthe roller and the groundare = 0.25 and = 0.2 .
Find:
(a) The maximum force P that can be applied to the handleso that the roller rolls on the ground without slipping(b) The angular acceleration and acceleration of the
roller with this P(c) The angular acceleration and acceleration of the
roller with P = 77 lb f .
Review Problems
8/12/2019 MT 2 Review
18/20
Example 5 (continued)Solution:
Equations of Motion: Themass moment of inertia of theroller about its mass center is
= : cos 30 = 03 .
(1)+
= 1 = 1 03 .
1.5 = 1.7469 slugft 2. Therefore,
+ = : sin 30 5 0 = 0 (2)+ = : 1.5 = 1.7469 (3)
Since the roller is required to be on the verge of slipping,= = 1.5 (4)
= = 0. 25 (5)
(a) and (b)
Review Problems
8/12/2019 MT 2 Review
19/20
Example 5 (continued)
Solving Eqs. (1) through (5) yields= 18.93 rad/s 2 = 76.37 lb f= 88.5 lb f = 28.39 ft/s 2 = 22.05 lb f
(c) EOMs
Since P > 76.37 , the roller is slipping. If we set P = 77 lb fand = = 0.2 in Eqs. (1) and (3), we have thefollowing EOMs:77 cos 30 0.2 =
03 .
(1) 77sin30 5 0 = 0 = 88.5 (2)
0.2 1.5 = 1.7469 (3)
Review Problems
8/12/2019 MT 2 Review
20/20
Example 5 (continued)(c) continuedUsing known values and solving Eqs. (1) through (3)yields
= 15.19 rad/s 2 = 77.0 lb f= 88.5 lb f = 31.54 ft/s 2 = 17.7 lb f
Comparing these results to the no-slip results tells us thatslipping causes the rollers translational acceleration toincrease and its rotational acceleration to decrease which isintuitively correct since the frictional force, resisting thetranslational motion and causing the rotation, is reduced.
Review Problems