P0320 -

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

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

Unformatted text preview: 3.20. CHAPTER 3, PROBLEM 20 175 3.20 Chapter 3, Problem 20 Problem: A small sphere of mass m has a tangential speed V as it arrives at the top of a frictionless cylindrical surface of radius R. At time t = 0, the sphere continues sliding along the surface. At an angle θ = θs where the normal reaction force between the sphere and cylinder, N , is zero, the sphere separates from the surface. (a) State the equations governing the motion up to the separation point in cylindrical coordinates. Let g denote the acceleration of gravity. ˙ ˙¨ ˙ (b) Multiply the equation for the circumferential motion by θ and, noting that θθ = d(θ2 /2)/dt, verify that V2 2g ˙ θ2 = 2 + (1 − cos θ) R R ˙ (c) Using the result for θ2 from Part (b), solve for N as a function of m, V , g and cos θ. (d) If the separation angle is θs = cos−1 ( 5 ), what is V 2 /(gR)? 6 .. . . ... . ............... ... ........... .. . . . .. .. ....... . . . ............. ..... . ........................... .. . . ......... .................... . . .......................................................... . .............................................................. . . .. ................................................................. . . .............................................. .. ................... ................................................................ .................................................................................................... . . . .. ..................................................................................................... . . .. .. .. .. ........... .... ... . ............................................................. . ......... . .............................................................................................. . . ..................... . .. .................................................................................................... . . ..................................................................................................... . . .. ...................................................................................................... . .. . .......................... . . .. . . .. ... . .................................................................................................................................... . . .. .................................................................................................................................... ... ........................................................................................................... ... . ............ ....... . .. . . . .. .. . . ................................................ . .. .................................................. ............................... .. .... ............................................................................................................ .. .................................................................................................................... . .. ..................................................................................................................... ....... ............ . ....................................................................................... ......................................................................................................................... . .. ................................................................................................................................................... ........................................................................................................................................................ ............... ............................................................................. . . . ... .. ............................................................. ...................................................................................................................................................................................................................................................... ............................................................................................................................ ... ......................................................................................................................................................................................................................................................... .. . . . .................................. ................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... g . . . . . . . . . . . . . . . . . ... ... . ... .. .. . . vV θ v R R v Solution: Note that, with θ defined as indicated in the figure, positive velocities and forces are in the clockwise direction. (a) Newton’s Second Law tells us that ˙ Fr = m r − rθ2 ¨ and ¨ ˙˙ Fθ = m rθ + 2rθ Up to the point of separation, the ball moves on the cylindrical surface, wherefore r = R and r = r = 0. ˙¨ Therefore, ˙ ¨ and Fθ = mRθ Fr = −mRθ2 There are two forces acting on the sphere, viz., the ball’s weight, mg , and the normal reaction force, N . As shown below, mg acts vertically downward and N acts radially outward. N mg v θ mg s n θ mg cos θ ......................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... Thus he equa ons govern ng he mo on up o he separa on po n are N − mg cos θ = −mRθ2 =⇒ and ¨ mg s n θ = mRθ (b) Now mu p y ng he c rcumferen a equa on hrough by θ we f nd ¨ mg θ s n θ = mRθθ g d 1d (− cos θ) = R θ2 dt 2 dt 176 Integrating once yields −g cos θ = CHAPTER 3. FORCE AND ACCELERATION 1 ˙2 Rθ + constant 2 ˙ But, θ = V /R when the sphere is at the top of the cylinder where θ = 0. Thus, −g = ˙ Solving for θ2 , we find V 2 2g ˙ θ2 = 2 + (1 − cos θ) R R ˙ (c) Substituting the result for θ2 from Part (b) into the radial equation of motion, we have N − mg cos θ = −mR Thus, solving for N yields V2 + mg (3 cos θ − 2) R (d) When the normal reaction force vanishes, the sphere has reached the angle, θs , at which it separates from the cylinder. Setting N = 0 in the equation derived in Part (c), there follows N = −m mg (3 cos θ − 2) = m Thus, the separation angle follows from 3 cos θs = 2 + We are given θs = cos−1 ( 5 ). Hence, 6 3· Thus, simplifying yields V2 1 = gR 2 V2 5 =2+ 6 gR =⇒ 5 V2 =2+ 2 gR V2 gR V2 R V2 2g V2 + (1 − cos θ) = −m − 2mg (1 − cos θ) R2 R R 1 V2 + constant R 2 R2 =⇒ − g cos θ = 1 ˙2 1 V 2 Rθ − −g 2 2R ...
View Full Document

This note was uploaded on 10/15/2010 for the course AME 301 at USC.

Ask a homework question - tutors are online