p0710

# p0710 - v B = ω R j Ω R w − 2 j − 1 2 i W = − 1 2...

This preview shows page 1. Sign up to view the full content.

7.10. CHAPTER 7, PROBLEM 10 315 7.10 Chapter 7, Problem 10 Problem: Ad i sko fr ad iu s R is mounted on L-shaped Rod CD and rotates with constant angular velocity ω as shown. Rod CD rotates with constant angular velocity about the z axis. Determine the absolute velocity and acceleration of Point B, which is the lowest point on the disk relative to the xy plane. Solution: An observer in a coordinate frame rotating with angular velocity = k about Point D sees the motion of Point B as pure rotation with angular velocity ω = ω i about the center of the disk (Point C). Thus, the relative velocity and acceleration vectors are v I B = ω × r B/C = ω i × ( R k )= ω R j a I B = ω × v I B = ω i × ω R j = ω 2 R k The absolute velocity of Point B is given by v B = v I B + × r B/D where r B/D is given by r B/D = 2 R i + 1 2 R j R
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: v B = ω R j + Ω R w − 2 j − 1 2 i W = − 1 2 Ω R i + ( ω − 2 Ω ) R j Point B’s absolute acceleration is a B = a I B + ˙ Ω × r B/D + 2 Ω × v I B + Ω × ( Ω × r B/D ) where v I B is the collar’s velocity as seen by the rotating observer and r B/D is its position vector relative to the collar. Now, we know that ˙ Ω = , while v I B and a I B are as given above. Thus, the absolute acceleration of Point B is a B = ω 2 R k + 2 Ω k × ω R j + Ω k × } Ω k × w − 2 R i + 1 2 R j − R k W] = ω 2 R k − 2 Ω ω R i + Ω k × } Ω R w − 2 j − 1 2 i W] = ω 2 R k − 2 Ω ω R i + Ω 2 R } 2 i − 1 2 j ] = 2( Ω − ω ) Ω R i − 1 2 Ω 2 R j + ω 2 R k...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online