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Today’s Objectives : Students will be able to: a) Resolve the acceleration of a point on a body into components of translation and rotation.

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RELATIVE MOTION ANALYSIS: ACCELERATION (Section 16.7)

Today’s Objectives:

Students will be able to:

a) Resolve the acceleration of a point on a body into components of translation and rotation.

b) Determine the acceleration of a point on a body by using a relative acceleration analysis.

In-Class Activities:

•Check homework, if any

•Reading quiz

•Applications

•Translation and rotation components of acceleration

•Relative acceleration analysis

•Roll-without-slip motion

•Concept quiz

•Group problem solving

•Attention quiz

READING QUIZ

1. If two bodies contact one another without slipping, and the points in contact move along different paths, the tangential components of acceleration will be ______ and the normal components of acceleration will be __________.

A) the same, the same B) the same, different

C) different, the same D) different, different

2. When considering a point on a rigid body in general plane motion,

A)it’s total acceleration consists of both absolute acceleration and relative acceleration components.

B)it’s total acceleration consists of only absolute acceleration components.

C)it’s relative acceleration component is always normal to the path.

D)None of the above.

APPLICATIONS

In the mechanism for a window, link AC rotates about a fixed axis through C, while point B slides in a straight track. The components of acceleration of these points can be inferred since their motions are known.

To prevent damage to the window, the accelerations of the links must be limited. How can we determine the accelerations of the links in the mechanism?

APPLICATIONS (continued)

The forces delivered to the crankshaft, and the angular acceleration of the crankshaft, depend on the speed and acceleration of the piston in an automotive engine.

How can we relate the accelerations of the piston, connection rod, and crankshaft in this engine?

=

+

dv

dv

dv

/

B

A

B

A

dt

dt

dt

These are absolute accelerations of points A and B. They are measured from a set of fixed x,y axes.

This term is the acceleration of B with respect to A.

It will develop tangential and normal components.

RELATIVE MOTION ANALYSIS: ACCELERATION

The equation relating the accelerations of two points on the body is determined by differentiating the velocity equation with respect to time.

The result isaB= aA+ (aB/A)t+ (aB/A)n

Graphically:

aB = aA+ (aB/A)t + (aB/A)n

=

+

RELATIVE MOTION ANALYSIS: ACCELERATION

The relative tangential acceleration component (aB/A)t is ( x rB/A) and perpendicular torB/A.

The relative normal acceleration component (aB/A)n is (-2rB/A) and the direction is always from B towards A.

RELATIVE MOTION ANALYSIS: ACCELERATION (continued)

Since the relative acceleration components can be expressed as (aB/A)t = rB/A and (aB/A)n = - 2rB/A the relative acceleration equation becomes

aB= aA + rB/A - 2rB/A

Note that the last term in the relative acceleration equation is not a cross product. It is the product of a scalar (square of the magnitude of angular velocity, w2) and the relative position vector, rB/A.

APPLICATION OF RELATIVE ACCELERATION EQUATION

In applying the relative acceleration equation, the two points used in the analysis (A and B) should generally be selected as points which have a known motion, such as pin connections with other bodies.

In this mechanism, point B is known to travel along a circular path, so aB can be expressed in terms of its normal and tangential components. Note that point B on link BC will have the same acceleration as point B on link AB.

Point C, connecting link BC and the piston, moves along a straight-line path. Hence, aC is directed horizontally.

PROCEDURE FOR ANALYSIS: RELATIVE ACCELERATION ANALYSIS

1.Establish a fixed coordinate system.

2.Draw the kinematic diagram of the body.

3.Indicate on it aA, aB, , , and rB/A. If the points A and B move along curved paths, then their accelerations should be indicated in terms of their tangential and normal components, i.e., aA= (aA)t + (aA)n and aB = (aB)t + (aB)n.

4.Apply the relative acceleration equation:

aB= aA+ rB/A- 2rB/A

5.If the solution yields a negative answer for an unknown magnitude, it indicates the sense of direction of the vector is opposite to that shown on the diagram.

EXAMPLE 1

Given:Point A on rod AB has an acceleration of 3 m/s2 and a velocity of 2 m/s at the instant the rod becomes horizontal.

Find:The angular acceleration of the rod at this instant.

Plan:Follow the problem solving procedure!

Solution:First, we need to find the angular velocity of the rod at this instant. Locating the instant center (IC) for rod AB (which lies above the midpoint of the rod), we can determine :

wA = vA/rA/IC = vA/(5/cos 45) = 0.283 rad/s

EXAMPLE 1 (continued)

Since points A and B both move along straight-line paths,

aA= 3 (cos 45 i - sin 45 j) m/s

aB = aB(cos 45 i + sin 45 j) m/s

Applying the relative acceleration equation

aB = aA + a x rB/A – w2rB/A

(aB cos 45 i + aB sin 45 j = (3 cos 45 i – 3 sin 45 j)

+ (ak x 10 i) – (0.283)2(10i)

EXAMPLE 1 (continued)

By comparing the i, j components;

aB cos 45 = 3 cos 45 – (0.283)2((10)

aB sin 45 = -3 sin 45 + a(10)

Solving:

aB = 1.87 m/s2

= 0.344 rad/s2

BODIES IN CONTACT

Consider two bodies in contact with one another without slipping, where the points in contact move along different paths.

In this case, the tangential components of acceleration will be the same, i. e.,

(aA)t = (aA’)t (which implies aBrB = aCrC ).

The normal components of acceleration will not be the same.

(aA)n (aA’)n so aAaA’

EXAMPLE: ROLLING MOTION

Another common type of problem encountered in dynamics involves rolling motion without slip; e.g., a ball or disk rolling along a flat surface without slipping. This problem can be analyzed using relative velocity and acceleration equations.

As the cylinder rolls, point G (center) moves along a straight line, while point A, on the rim of the cylinder, moves along a curved path called a cycloid. If w and a are known, the relative velocity and acceleration equations can be applied to these points, at the instant A is in contact with the ground.

EXAMPLE: ROLLING MOTION(continued)

•Velocity analysis.

Since no slip occurs, vA = 0 when A is in contact with ground. From the kinematic diagram:vG = vA + w x rG/A

vG i = 0 + (-w k) x (r j)

vG = wr or vG = wr i

•Acceleration.

Since G moves along a straight-line path, aG is horizontal. Just before A touches ground, its velocity is directed downward, and just after contact, its velocity is directed upward. Thus, point A accelerates upward as it leaves the ground.

aG = aA + a x rG/A – w2rG/A => aGi = aAj + (-ak) x (r j) – w2(r j)

Evaluating and equating i and j components:

aG = ar and aA = w2r or aG = ar i and aA = w2r j

These results can be applied to any problem involving roll without slip.

EXAMPLE 2

Given:The ball rolls without slipping.

Find:The accelerations of points A and B at this instant.

Plan:Follow the solution procedure.

Solution:Since the ball is rolling without slip, aOis directed to the left with a magnitude of

aO= r= (4 rad/s2)(0.5 ft)=2 ft/s2

EXAMPLE 2 (continued)

Now, apply the relative acceleration equation between points O and B.

aB = aO + a x rB/O – w2rB/O

aB = -2i + (4k) x (0.5i) – (6)2(0.5i)

= (-20i + 2j) ft/s2

Now do the same for point A.

aA = aO + a x rA/O – w2rA/O

aA = -2i + (4k) x (0.5i) – (6)2(0.5j)

= (-4i – 18j) ft/s2

1.If a ball rolls without slipping, select the tangential and normal components of the relative acceleration of point A with respect to G.

A) r i+ 2r jB) -r i+ 2r j

C) 2r i - r jD) Zero.

B

y

x

CONCEPT QUIZ

2.What are the tangential and normal components of the relative acceleration of point B with respect to G.

A)-2r i- r jB) -r i+ 2r j

C) 2r i- r jD) Zero.

GROUP PROBLEM SOLVING

Given: The disk is rotating with

= 3 rad/s, = 8 rad/s2 at this instant.

Find:The acceleration at point B, and the angular velocity and acceleration of link AB.

Plan:Follow the solution procedure.

Solution:

At the instant shown, points A and B are both moving horizontally. Therefore, link AB is translating, meaning wAB = 0.

(aA)t

AB

(aA)n

aB = aA + aAB x rB/A – w2ABrB/A

Where

aAn = 0.2*32 = 1.8 m/s2

aAt = 0.2*8 = 1.6 m/s2

rB/A = 0.4 cos 30i – 0.4 sin 30j

aAB = aABk, wAB = 0

GROUP PROBLEM SOLVING (continued)

Draw the kinematic diagram and then apply the relative-acceleration equation:

aBi = (1.6i – 1.8j) + aABk x (0.4 cos 30i – 0.4 sin 30j)

= (1.6 + 0.4 sin 30 aAB)i + (-1.8 + 0.4 cos 30 aAB)j

aAt

AB

aAn

Solving:

aB = 2.64 m/s2

aAB = 5.20 rad/s2

GROUP PROBLEM SOLVING (continued)

By comparing i, j components:

aB = 1.6 + 0.4 sin 30 aAB

0 = -1.8 + 0.4 cos 30 aAB

2 ft

1 ft

ATTENTION QUIZ

- Two bodies contact one another without slipping. If the tangential component of the acceleration of point A on gear B is 100 ft/sec2, determine the tangential component of the acceleration of point A’ on gear C.
- A)50 ft/sec2 B) 100 ft/sec2
- C) 200 ft/sec2 D) None of above.

2.If the tangential component of the acceleration of point A on gear B is 100 ft/sec2, determine the angular acceleration of gear B.

A)50 rad/sec2 B) 100 rad/sec2

C) 200 rad/sec2 D) None of above.

End of the Lecture

Let Learning Continue