p0254 - 160 CHAPTER 2. PARTICLE KINEMATICS 2.54 Chapter 2,...

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Unformatted text preview: 160 CHAPTER 2. PARTICLE KINEMATICS 2.54 Chapter 2, Problem 54 Problem: A girl riding on a ferris wheel throws a ball with initial speed vo = − 1 ΩR i + vo k, relative 2 to the rotating wheel. She catches the ball at a time when her seat is at an angle θ to the horizontal. The ferris wheel has radius R and constant angular rotation rate Ω. Ignore effects of friction on the ball’s motion. (a) Set up an equation from which θ can be determined. Your equation’s coefficients should be in terms of vo , Ω, R and gravitational acceleration, g . It is a transcendental equation involving both sin θ and cos θ that cannot be solved in closed form. HINT: Solve first for t as a function of θ and Ω from the x-direction equation, and then substitute the result into the z -direction equation. (b) Assuming the girl catches the ball when θ = 0o , determine vo as a function of Ω, R and g . . . . .. . . . . . . . . . . . .. . . . . . . . . . . . .. .................... .. ... ...................... .................................. .................................. .. ... ...... . . . .. . . .... .. . ... . .. ... ... . .......... ......... ......... . .... . . ............. .. . . ...... . . .. .... .. ..... .... . ...... . ... ....... ... . ... . . . .... . ..... . ..... . .. . .. ..... . ..... .. . .. .... . . .. . .. .... ...... . ..... ..... .. . ... ..... . ... . . . ... . . . . .... . . .. .................... . .... .... . . . ..... . .. .... .. ................. . . . .... ..... ... .. ..... .... ..................... . . ...................... ................... ... . . . ... . . .. . ... .. .................. ... .... . . ............. .. . ................... ... ..... . ... .. . ......... ....... . . . .................... ... . .. . .. ... . . . . ... . . ... ... ... .. .. . . ... ... . . .. . ... .. .. . . . . ... . ... ... .. . . .. ... . ... . . .. .. ... ... ... ... .. .. ... .. .. ... ... .. .. .. ... ... . ... ... .. ... . .. . ... . ... . ... .. .. . ... . . .. .. ... . ... ... . .... .. ... .. ... . ... .. . . . .. .. ... .. .. . . .. .. ... .. ... .. ... . ... .. . .. . .. .. . .. .. . .... .. . ... ... . .. . .. . ... ... . .. . . .. . .... . .. .. .. .. . .. ... ......... .. . . .... . ... ......... ... ... .. .. . .. . . .. .... .. . . .. ... . . ...... . .. . . . . .. . . .... ... ....... . .. .. ...... . ... .. . .. .. . .. . .. . . .. ... . . ... . . . .. . . ............ .. ... .. .... .. . .. ....... . . . .. . .. .. . ... .. ... . . . .. . . . . ... ..... ......... ... ... .. . .. . .. . ...... . . . .. . .. ....... ...... . . . . .. . .. ... . . . . ...................... .. . .. . .... . .... . .. . . .................... ................. .. ................. . . . . .... . . . . .. . . ................. .. . .. . . . ..... ...... .. .... . .. . .. . .. . . . . .. .. . .. . . . .. .. . ...... .. .. .. .. ...... . . .. .. . . .... ... . ...... . . .. .. . . .... .. . ...... . . . . . .. .. .. .. . ...... ... .. . ... .. .. ...... . .. . . .. .. . . ... ............ . .. .. .. . ... ..... ......... .. ... . .. . ..... .. . .. ... .. ... ... . . .. .. . ... . . .. .. .. ... . .. .. . ... ... ... .. .. .. . .. . .. . ... . .. ... .. .. ... .. . .. ... .. .. . ... . . ... ... . ... ... . .. ... . .. . ... ... . .. . .. ... .. ... . ... .. ... . .. . ... ... .. ... ... ... ... .. .. . ... .. .. ... .. . ... ... ... ... . .. ... ... . . . . .... . .... ... ... ...... ... .... . ...... ... ... .... . ..... ... ... .... . .... .. .. ... ...... .... . .... .... . .... ....... .. .... ..... . ... . .. .. ..... ..... . .. . ..... .... ..... . ... .... . .... . .. ..... .... ... .. .. ..... ... .... ..... ..... . ... . ..... .... . ..... ...... . o ... . ..... ...... .. . .. ..... .. . ....... ...... ..... .. ....... .. . .. . ....... ........ .. .. . ... .... . .......... ........ .......... . ......... . .................................................................................. . ............................................................................. . ....... ....... ....... .. .......................... ........ . .... . . .. ... ... ........... . .. . . .. . . . ..... .. ........... ..... ........................ ......................... . . .. ..................... . . ... . .. . ... ........... ................... z • Ω θ • R g = −g k v • x Solution: Ignoring frictional forces, the only force acting on the ball is gravity. Hence, the horizontal (x) component of its velocity is constant while the vertical (z ) component decreases due to gravitational acceleration. (a) Noting that the velocity of the ferris wheel at the point where the ball is launched is ΩR in the x direction, the absolute velocity components of the ball (i.e., relative to a stationary reference frame) are 1 1 vx = ΩR − ΩR = ΩR 2 2 and vz = vo − gt where t = 0 is the time at which the girl throws the ball. Integrating over time and using the fact that the ball’s coordinates at t = 0 are xB (0) = zB (0) = 0, the ball’s position relative to the fixed xz frame is given by 1 1 xB (t) = ΩRt and zB (t) = vo t − gt2 2 2 Note that the angle θ (in radians) and time t are related by θ = Ωt − π 2 2.54. CHAPTER 2, PROBLEM 54 Turning to the ferris-wheel seat’s position, we have xA (t) = R cos θ In general, the relative position of the ball is rB/A = rB − rA and zA (t) = R(1 + sin θ) 161 For the problem at hand, we seek the solution for which rB/A = 0. Hence, we conclude that for the x direction, 1 2 cos θ ΩRt = R cos θ =⇒ t= 2 Ω Similarly, for the z direction, 1 vo t − gt2 = R(1 + sin θ) 2 Substituting for t, we arrive at the desired equation for determining θ, viz., 2g 2vo cos θ − 2 cos2 θ = R(1 + sin θ) Ω Ω (b) Now, when θ = 0o , the equation for θ derived in Part (a) simplifies to 2vo 2g − 2 =R Ω Ω Solving for vo yields vo = 1 g ΩR + 2 Ω ...
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This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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