# Exam 3 - quiz 03 – BAUTISTA ALDO – Due Apr 5 2006 9:00...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: quiz 03 – BAUTISTA, ALDO – Due: Apr 5 2006, 9:00 pm 1 Version number encoded for clicker entry: V1:1, V2:1, V3:5, V4:5, V5:1. Question 1 Part 1 of 1. 10 points. A uniform bar of length L and weight W is attached to a wall with a hinge that exerts on the bar a horizontal force H x and a vertical force H y . The bar is held by a cord that makes a 90 ◦ angle with respect to bar and angle θ with respect to wall. The acceleration of gravity g = 9 . 8 m / s 2 . L 90 ◦ W θ What is the magnitude of the horizontal force H x on the pivot? 1. H x = 1 2 W tan θ 2. H x = 1 2 W cos θ 3. H x = 1 2 W sin 2 θ 4. H x = 1 2 W sin θ cos θ correct 5. H x = 1 2 W sin θ 6. H x = 1 2 W cos 2 θ Explanation: Analyzing the torques on the bar, with the hinge at the axis of rotation, we have X τ = L T- L 2 sin θ W = 0 , so, T = 1 2 W sin θ . Analyzing the force on the bar, we have X F x = H x- T cos θ = 0 . Put T into this equation and get H x = 1 2 W sin θ cos θ . Question 2 Part 1 of 1. 10 points. A massless rod of length L has a mass 2 m fastened at its center and another mass m fastened at one end. On the opposite end from the mass m , the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity ω max as the rod swings through its lowest (vertical) posi- tion? 1. ω max = r 2 g L 2. ω max = r 3 g 2 L 3. ω max = r g L 4. ω max = r 12 g 5 L 5. ω max = r 8 g 3 L correct 6. ω max = r 4 g 5 L Explanation: The mechanical energy of the system is con- served. Measuring heights from the point at quiz 03 – BAUTISTA, ALDO – Due: Apr 5 2006, 9:00 pm 2 the bottom of the rod when it is vertical, the initial potential energy of the system is U i = (3 m ) g L . The potential energy at the bottom of the swing, U f = 2 m g ( L/ 2) = m g L . To calculate the final kinetic energy, we need the moment of inertia of the system, I = 2 m ( L/ 2) 2 + m L 2 = 3 m L 2 / 2 . There- fore, 3 m g L = m g + 1 2 3 2 m L 2 ω 2 , ω max = r 8 g 3 L . Question 3 Part 1 of 2. 10 points. A student sits on a rotating stool holding two 3 . 8 kg masses. When his arms are ex- tended horizontally, the masses are 0 . 78 m from the axis of rotation, and he rotates with an angular velocity of 2 . 4 rad / sec. The stu- dent then pulls the weights horizontally to a shorter distance 0 . 24 m from the rotation axis and his angular velocity increases to ω 2 . ω i ω f For simplicity, assume the student himself plus the stool he sits on have constant com- bined moment of inertia I s = 4 . 8 kg m 2 . Find the new angular velocity ω 2 of the student after he has pulled in the weights. 1. 3 . 93273 rad / s 2. 4 . 06179 rad / s 3. 4 . 18764 rad / s 4. 4 . 31811 rad / s correct 5. 4 . 45277 rad / s 6. 4 . 59392 rad / s Explanation: Let : M = 3 . 8 kg , R 1 = 0 . 78 m , ω 1 = 2 . 4 rad / sec , R 2 = 0 . 24 m , and ω 2 = 4 . 31811 rad / sec , As the student moves his arms, his moment of inertia changes from...
View Full Document

## This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

### Page1 / 14

Exam 3 - quiz 03 – BAUTISTA ALDO – Due Apr 5 2006 9:00...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online