ENME361 Quiz 4 Solution

ENME361 Quiz 4 Solution - B. Balachandran ENME 361 Fall...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
B. Balachandran ENME 361 Fall 2008 Solution to Quiz #4 (October 9, 2008; Duration: 15 minutes) 1) Given a vector velocity v of the form ( ) cos sin xb b θ θθ =+ + ±± ± vi j where x = x ( t ), = ( t ), and b is a constant, construct () 1 2 T = vv and determine an expression for dT dt ⎛⎞ ⎜⎟ ⎝⎠ ± Solution : 22 2 11 () ( 2 c o s ) Tx b x b == + + v.v 2 cos T bx b ± ± ± 2 cos sin b b x dt ± ± ± 2) Consider vertical oscillations and determine the kinetic energy for the system shown in the figure below, where an unbalance of mass m o rotates with an angular speed ω , and this mass is located a fixed distance ε from the center of rotation O . Note that M does not include the unbalance m o . M m o k c t X Y i j O
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Solution : To determine the kinetic energy associated with the unbalanced mass, we first write the position vector from point O’ , the origin of the inertial reference frame, to mass m o as o m r( c o s ) ( s i n ) ht yt ε ωε ω =+ ++ i j where h is the horizontal offset of point O from point O’ and y is measured in the vertical direction from the static-equilibrium position. Therefore, the absolute velocity of the unbalanced
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

ENME361 Quiz 4 Solution - B. Balachandran ENME 361 Fall...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online