{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework5.20100217.4b7cb31d74c6b5.44352284

# homework5.20100217.4b7cb31d74c6b5.44352284 - D 2 r = ¨ x e...

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

TAM 412, Homework 5, due February 24, 2010 Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign [email protected] 1. Do exercise 1.9 of the textbook. 2. Do exercise 1.13 of the textbook. 3. Do exercise 1.14 of the textbook. To this end, note the following fundamental result (cf. Example 1.21), which hopefully clari fi es the sign confusion in class on 2/17/10. The latter was due to a mixup of whether the tilde referred to the inertial or rotating reference frame. Let R = ( T , O ) denote a reference frame which rotates relative to the inertial reference frame ˜ R = ³ ˜ T , O ´ with angular velocity ω . Let r denote the position vector from O to a particle. Then, D r = ˙ x e x + ˙ y e y + ˙ z e z is the velocity relative to the rotating reference frame and ˜ D r = ˙ ˜ x e ˜ x + ˙ ˜ y e ˜ y + ˙ ˜ z e ˜ z is the velocity relative to the inertial reference frame. Similarly,
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: D 2 r = ¨ x e x + ¨ y e y + ¨ z e z is the acceleration relative to the rotating reference frame and ˜ D 2 r = ¨ ˜ x e ˜ x + ¨ ˜ y e ˜ y + ¨ ˜ z e ˜ z is the acceleration relative to the inertial reference frame. It follows that ˜ D r = D r + ω × r and ˜ D 2 r = D 2 r + ( D ω ) × r + 2 ω × ( D r ) + ω × ( ω × r ) In Example 1.21, the particle is acted upon by a force N + m g and D ω = ˜ D ω = . It follows that N + m g = m ˜ D 2 r = mD 2 r + 2 m ω × ( D r ) + m ω × ( ω × r ) . 4. Do exercise 1.15 of the textbook. 5. Do exercise 1.18 of the textbook. 6. Do exercise 1.19 of the textbook. 7. Identify and discuss (using at least 150 words per topic) the four most important points in Chapter 1. 1...
View Full Document

{[ snackBarMessage ]}