Homework 12-solutions - yindeemark (rry82) Homework 12...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: yindeemark (rry82) Homework 12 chelikowsky (59005) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Assume: A bullet of mass m and cube of mass M undergo an inelastic collision, where m M . Note: The moment of inertia of this cube (with edges of length 2 a and mass M ) about an axis along one of its edges is 8 M a 2 3 . A solid cube is resting on a horizontal sur- face. The cube is constrained to rotate about an axis at its bottom right edge (due to a small obstacle on the table). A bullet with speed v min is shot at the left-hand face at a height of 4 3 a . The bullet gets embedded in the cube. 2 a M mvectorv min 4 3 a M g Find the minimum value of v min required to tip the cube so that it falls its right-hand face. 1. v min = m M radicalbigg 5 g a parenleftBig 2 1 parenrightBig 2. v min = M m radicalbigg 3 g a parenleftBig 2 1 parenrightBig correct 3. v min = m M radicalbigg 2 g a parenleftBig 2 1 parenrightBig 4. v min = m M radicalbigg 3 g a parenleftBig 5 1 parenrightBig 5. v min = m M radicalbigg 3 g a parenleftBig 3 1 parenrightBig Explanation: Basic Concepts: summationdisplay vector L = const U + K = 0 For the cube to tip over the center of mass (CM) must rise so that it is over the axis of rotation AB . To do this the CM must be raised a distance of a parenleftBig 2 1 parenrightBig . From conservation of energy M g a parenleftBig 2 1 parenrightBig = 1 2 I cube 2 . (1) From conservation of angular momentum 4 a 3 mv = parenleftbigg 8 M a 2 3 parenrightbigg = parenleftBig mv 2 M a parenrightBig . (2) Thus, substituting Eq. 2 into 1, we have 1 2 parenleftbigg 8 M a 2 3 parenrightbiggparenleftbigg m 2 v 2 4 M 2 a 2 parenrightbigg = M g a parenleftBig 2 1 parenrightBig Solving for v yields v min = M m radicalbigg 3 g a parenleftBig 2 1 parenrightBig . 002 10.0 points If you walk along the top of a fence, why does holding your arms out help you to balance? 1. Your momentum is decreased. 2. Your rotational inertia is decreased. 3. Your rotational inertia is increased. cor- rect 4. Your momentum is increased. Explanation: Your rotational inertia increases when your arms are outstretched, which increases the resistance to a change in your rotational state. 003 10.0 points The angle of the inclination is 34 . 9 , the outer part of the large pulley has a radius of 3 r and the inner part of the large pulley has a radius of r . yindeemark (rry82) Homework 12 chelikowsky (59005) 2 3 r r /2 r m T Find the mass m needed to balance the 820 kg truck on the incline. The acceleration of gravity is 9 . 8 m / s 2 . Assume all pulleys are frictionless and massless and the cords are massless....
View Full Document

This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

Page1 / 8

Homework 12-solutions - yindeemark (rry82) Homework 12...

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

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