p0320 - 230 CHAPTER 3. FORCE AND ACCELERATION 3.20 Chapter...

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Unformatted text preview: 230 CHAPTER 3. FORCE AND ACCELERATION 3.20 Chapter 3, Problem 20 Problem: A girl is swinging on a swing. The combined mass of the girl and the swing is m. At the position shown, the tension in each of the two cables supporting the swing is T . The length of each cable is L and they are at an angle θ to the horizontal. Gravitational acceleration is g . We wish to determine the girl’s acceleration and velocity. (a) Using natural coordinates, state the governing equations. Determine the girl’s velocity vector (as a function of L, T , m, θ and g ) and acceleration vector (as a function of v , dv/dt and L) in terms of conventional natural-coordinate unit vectors, n and t. (b) Using cylindrical coordinates, state the governing equations. Determine the girl’s velocity vector (as a function of L, T , m, θ and g ) and acceleration vector (as a function of v , dv/dt and L) in terms of conventional cylindrical-coordinate unit vectors, er and eθ . ... ... . ..... .. . . ............. .. .. ............................................................................................................ .. ..... . . . . . . .... .... . . . . . .. . . . . . . ................................................................................................................ ........................... ...................................................................... ................................................................................................................ ................. ...... ......... .................... .. ............................. . ....................................................................................................................... . .... . ....................................................................................... . . . ................................................................................. .. . . . ........... . . . . . . ........................... ......................................................................................... . .. . . . . . . . .... . .. . .... . .. .. .. .... .. ... . ... . . . ... . . .. ... . .. .... ... ...... . . . .. . . . ... .. . .. .. . . ... ..... . . ... ..... . . . . ... ... .. . ... ... ... . . . . ... ... ... ... . . . ... ... .. ... ... ... ... .. . . .. . ... ... . ... ... . ... ... . ...... ... ... ... . ... .. .. .. . . ....... ........ .. .... .... ... ... ... ... ... ... ... ... ... ... ... ..... .... ... . θ • L 2T g v Solution: For each part of this problem, the first thing we must do is establish the orientation of the unit vectors. Then, we can use Newton’s Second Law to determine the acceleration and velocity vectors. (a) For natural coordinates, unit vector t is tangent to the path of motion (a semicircle here) and points in the direction of the velocity vector. The unit vector n points toward the center of rotation. The following figure shows the unit vectors. ... ... .. .. ....................................................................................................................... . . .. . . . . . . . . . . ...... ... .. ..... . . . . . ..... .... .. .................................................................................................................. . .. . . . . ...... .. . . . .. . . .. .. . . ....................... ............................................................................................................... ........................................ ........... . . ...................................................................................................................... .. . . . . . . . . ......................... ........ .. .. ........... .. . . ............................................................................................................ . .. . .................................................................................. ............................... . .. ............................................................................. ................................. ................................ . . . . .. . .... . .. .. ..... . . . .. . . .. ... ... ... . ... . . .... .... . . .. . . . .. .. . . .... . .. .... . . . . . . . . . ... ... ... ..... . . . . . . . . . . ... . . ... .. . . . . . .. . . ... ... . ... ... . . . ... . . ... . . ... .. . ... .. . ... ... .. .. . ... .. . ... . . ... ... ... ... ... . .. . . .. .. . ... .. ..... . .. .. . .... . ... ........ .. ....... .. .. . . .. . ..... ... ... . ... .. .. ... .. ..... .. .. .... .. . ... . ...... ..... . ..... .. .... .. ... . .. ... .. .. ..... ... . . .. .. ... . . ... . . .. ... . . ... . .. .. . .. ... ... . . . . ... ..... . . . . . . . . . . . . . . . . . .... .............. ... . θ • L 2T g n t In terms of n and t, the forces acting are 2T = 2T n and Thus, Newton’s Second Law tells us that mg = −mg sin θ n + mg cos θ t dv dt v2 L Fti = mg cos θ = m i i Fni = 2T − mg sin θ = m 3.20. CHAPTER 3, PROBLEM 20 231 where we use the fact that, for natural coordinates, at = dv/dt and an = v 2 /L. Therefore, the acceleration vector is v2 dv a= n+ t L dt Also, the momentum equation in the normal direction permits solving for the girl’s speed, v . There follows v = vt where v= 2LT − gL sin θ m (b) For cylindrical coordinates, unit vector eθ is tangent to the path of motion. The unit vector er points radially outward from the center of rotation. The following figure shows the unit vectors. ... ... ... ... .............................................................................................. . ..... . . . . . . . . . . ..... . . . . . . . . . . . . . .. ..................................................................................................................... .. .. ................ .... . ............. .......... ........ ............................................................................................................ . . . . . . . . ....... . ... ......................... . ..... ..... . . ............................................................................. ..................................... . . . .. .. .. .. ............... ......... .. . . . . . . . . . . . ............ . . . ... .......... .. ......... .............................................................................................. . .. . .............................................................................................................................................. ................................................................................................................ ............................... . ... ....... . . . . . .... . . . .... .. . ... .. .. . . . . .. .. ... . .. . . . . ... . .. .. .... . . .. . . . . ... ...... . . . ... ...... . . .. . . . ... .. . . . ... ... . . . . . ... ... . . . . ... ... .. . . ... . ... .. . . . . ... .. ... . ... .. .. . . . .. . . ... ... . ... ... ... . . . .. . . . .. .. . .... . . .. . .... ... . ... . .. . . ......... .. . ... ........ . ... . .. ..... ... ... . ... .. .. .. .... .. ... ...... ... ....... .... .. .. . .. .. . .. ...... . .. ........ . . .. ... ..... . ... ...... . .. ..... . ... .. .. ... . . ... . . .. ... . . . ... ... . .. ... . .. ... .... . ... ..... .. . . . . . . . . . . . ... . . .... . .... . . . . . . . . . . . . . . . .. .. . .... r ..... . θ • L 2T g e eθ In terms of er and eθ , the forces acting are 2T = −2T er Thus, Newton’s Second Law tells us that ¨ Fθi = mg cos θ = maθ = m rθ + 2rθ ˙˙ i and mg = mg sin θ er + mg cos θ eθ i ˙ Fri = mg sin θ − 2T = mar = m r − rθ2 ¨ For this geometry, we have r = L, so that r = 0 and r = 0. Also, the velocity is purely circumferential so ˙ ¨ ˙ ˙ that v = Lθ, which tells us θ = v/L. Therefore, aθ = so that the acceleration vector is dv dt a=− and ar = − v2 L v2 dv er + eθ L dt Also, the momentum equation in the radial direction permits solving for the girl’s speed, v . There follows v = v eθ where v= 2LT − gL sin θ m ...
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