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Objectives : Describe the velocity of a rigid body in terms of translation and rotation components.PowerPoint Presentation

Objectives : Describe the velocity of a rigid body in terms of translation and rotation components.

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VELOCITY (Section 16.5)

- Objectives:
- Describe the velocity of a rigid body in terms of translation and rotation components.
- Perform a relative-motion velocity analysis of a point on the body.

As block A moves to the left with vA, it causes the link CB to rotate counterclockwise, thus vB is directed tangent to its circular path. Which link undergoes general plane motion? How to find its angular velocity?

Gear systems are used in many automobile automatic transmissions. By locking or releasing different gears, this system can operate the car at different speeds.

Disp. due to translation

drB = drA + drB/A

Disp. due to rotation

Disp. due to translation and rotation

RELATIVE MOTION ANALYSIS:

DISPLACEMENT

A body undergoes a combination of translation and rotation (general motion)

Point A is called the base point, has a known motion. The x’-y’ frame translates with the body, but does not rotate. The displacement of B:

+

RELATIVE MOTION ANALYSIS:

VELOCITY

The velocity at B is :(drB/dt) = (drA/dt) + (drB/A/dt)or

vB = vA + vB/A

Since the body is taken as rotating about A,

vB/A = drB/A/dt = w x rB/A

Here w will only have a k component since the axis of

rotation is perpendicular to the plane of translation.

VELOCITY (continued)

vB = vA + w x rB/A

When using the relative velocity equation, points A and B should generally be points on the body with a known motion. Often these points are pin connections in linkages.

Both points A and B have circular motion since the disk and link BC move in circular paths. The directions of vA and vB are tangent to the circular path of motion.

VELOCITY (continued)

vB = vA + w x rB/A

When a wheel rolls without slipping, point A has zero velocity.

Furthermore, point B at the center of the wheel moves along a horizontal path. Thus, vB has a known direction, e.g., parallel to the surface.

EXAMPLE 1 (Solution)

vB = vA + wAB x rB/A

vB i= -2 j + (wkx (0.2 sin 45 i - 0.2 cos 45 j ))

vB i= -2 j + 0.2 w sin 45 j + 0.2 w cos 45 i

Equating the i and j components gives:

vB = 0.2 w cos 45

0 = -2 + 0.2 w sin 45

Solving:

w = 14.1 rad/s or wAB= 14.1 rad/s k

vB = 2 m/s or vB = 2 m/s i

Given:Collar C is moving downward with a velocity of 2 m/s.

Find: The angular velocities of CB and AB at this instant.

EXAMPLE 2 (solution)

Link CB. Write the relative-velocity equation:

vB = vC + wCB x rB/C

vB i= -2 j + wCB k x (0.2 i - 0.2 j )

vB i = -2 j + 0.2wCB j + 0.2 wCB i

By comparing the i, j components:

i: vB = 0.2 wCB => vB = 2 m/s i

j: 0 = -2 + 0.2 wCB => wCB = 10 rad/s k

EXAMPLE 2 (continued)

Link AB experiences only rotation about A. Since vBis known, there is only one equation with one unknown to be found.

vB = wAB x rB/A

2 i = wAB k x (-0.2 j )

2 i = 0.2 wAB i

By comparing the i-components:

2 = 0.2 wAB

So, wAB = 10 rad/s k

Given: The crankshaft AB is rotating at 500 rad/s about a fixed axis passing through A.

Find: The speed of the piston P at the instant it is in the position shown.

GROUP PROBLEM SOLVING (solution)

1) First draw the kinematic diagram of link AB.

Applying the relative velocity equation

vB = -50 j m/s

GROUP PROBLEM SOLVING (continued)

2) Now consider link BC.

Applying the relative velocity equation:

vC = -50 j m/s

Let Learning Continue

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