{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 10

# Homework 10 - Bautista Aldo Homework 10 Due Nov 8 2005 4:00...

This preview shows pages 1–3. Sign up to view the full content.

Bautista, Aldo – Homework 10 – Due: Nov 8 2005, 4:00 am – Inst: Maxim Tsoi 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A uniform solid sphere I sphere = 2 5 m r 2 of mass 1 . 8 kg and radius r = 0 . 209 m, is placed on the inside surface of a hemispherical bowl of radius R = 1 . 6 m. The sphere is released from rest at an angle θ = 46 . 1 from the vertical and rolls without slipping (see the figure). The acceleration of gravity is 9 . 8 m / s 2 . θ R r What is the angular velocity of the sphere when it reaches the bottom of the bowl? Correct answer: 11 . 6914 rad / s. Explanation: Basic Concepts: Energy conservation (potential + kinetic = constant): U i + K i = U f + K f . Use energy conservation for this problem. The initial energy is the potential energy of the center of mass of the sphere above its lowest point. The distance between the center of mass of the sphere at its highest (rest) and its lowest points is h = ( R - r ) - ( R - r ) cos θ = ( R - r )(1 - cos θ ) = (1 . 6 m - 0 . 209 m)(1 - cos 46 . 1 ◦◦ ) = 0 . 426478 m so that U i = m g h = (1 . 8 kg) (9 . 8 m / s 2 ) (0 . 426478 m) = 7 . 52308 J , U f = 0 J , and Δ U = U i - U f = 7 . 52308 J . where m is the mass of the sphere. The moment of inertia of the sphere is I sphere = 2 5 m r 2 and ω = v r . The energy of the sphere at the bottom of the bowl is all kinetic. This final kinetic energy: K f = 1 2 m v 2 f + 1 2 I ω 2 f = 1 2 + 2 10 m v 2 f = 7 10 m v 2 f . Energy conservation gives 7 10 m v 2 f = m g h v f = r 10 7 g h = r 10 7 (9 . 8 m / s 2 ) (0 . 426478 m) = 2 . 4435 m / s . ω f = v f r = 2 . 4435 m / s 0 . 209 m = 11 . 6914 rad / s . 002 (part 1 of 3) 10 points A 3 kg bicycle wheel rotating at a 2676 rev / min angular velocity has its shaft supported on one side, as shown in the fig- ure. The wheel is a hoop of radius 0 . 5 m, and its shaft is horizontal. The distance from the center of the wheel to the pivot point is 0 . 5 m. When viewing from the left (from the posi- tive x -axes), one sees that the wheel is rotat- ing in a clockwise manner. Assume: All of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9 . 8 m / s 2 .

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

View Full Document
Bautista, Aldo – Homework 10 – Due: Nov 8 2005, 4:00 am – Inst: Maxim Tsoi 2 y z x b m g ω R The magnitude of the angular momentum of the wheel is given by 1. k ~ L k = 1 2 m R 2 ω 2. k ~ L k = 1 4 m R 2 ω 3. k ~ L k = 1 4 m R 2 ω 2 4. k ~ L k = m R 2 ω correct 5. k ~ L k = 1 2 m R 2 ω 2 6. k ~ L k = m R 2 ω 2 Explanation: Basic Concepts: ~ τ = d ~ L dt Solution: The magnitude of the angular momentum I of the wheel is L = I ω = m R 2 ω , and is along the negative x -axis. 003 (part 2 of 3) 10 points Let : m = 3 kg , ω = 2676 rev / min b = 0 . 5 m , and R = 0 . 5 m . Find the change in the precession angle after a 1 . 3 s time interval. Correct answer: 5 . 20964 . Explanation: Let : ω = 2676 rev / min = 2 π (2676 rev / min) (60 s / min) = 280 . 23 rad / s .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}