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EGR 2513.001 and 002

MT Exam #2Spring 2014

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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

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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 / .

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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 /

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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 /

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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)

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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

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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

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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.

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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

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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

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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.

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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.

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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)

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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 .

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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 .

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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)

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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)

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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.

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