p0320

# P0320 -

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

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