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Unformatted text preview: 138 CHAPTER 2. PARTICLE KINEMATICS 2.54 Chapter 2, Problem 54 Problem: Friction Disk A drives Friction Disk B with no slip occurring. Pulley C is attached to Friction Disk B and thus has the same rotation rate. A cable that wraps around the pulley supports Weight D. The angular velocity and angular acceleration of Friction Disk A are ωa and αa , respectively. Determine the linear velocity vector, vd , and linear acceleration vector, ad , for the weight. Express your answers as functions of d, ωa and αa . . . .. .......... ......... . . .. a ....... ... ............ . ........ ....... . ................................ .. .............. ... . ..................... ....................................................... .. ... ............. ...... ...................................................... .. .............................. . . . .. . ........................................... . ........................................... ............................................ . . . . ........................................... . .. . ........................................... ......................................... .. ................ . ........................................ ...................................... .............................. ................. . .......................... ................ ......... .. . .. . . . .. ............. . .. ................................ ....................................... ....................................... . .. .... ....... . . .................................. ... ................................... .............................................................. ............................................ ................................................. ... . . . . .. .......................................................................... ................................................................... ................................................................................. . . ........... ... ............................................................................. ................................................................................................ .............. .................................................................................. .................................................................................... ................................................ . .. ........... . .. ........ ............................................................ . ..................................................................................................... . . ........................................................................................ . .. . .. . .................................................................................................... .... .... ... .. .................. .......... . . ................................................................................................ . . . ..................................................................................... .. . . ...................... ........................................................... .. ........ ................... . .................................................................................. . ................................................................................... .. . ................................................. .... .............................................................................. ....................... .. .. ....................................................................... ................................................................. ... .... . .. .......................................................................... ........................................................ ... . ............... ................ .. . . . ....................................................................... ................................................................ .... ............................. .. ............................................................... .................................. ..................................................... ...... ............................................... . .. . .. .................................. . .. .. ... . ........................ . ................. . . . . . . . . . . . . . . . . . . ............................. . ............................. ............................. ............................. ............................. ............................. ............................. ............................. ............................. ............................. ............................. ....... ............................. ............................. . .. .. .. .. .. .. .. ω A • B ......... ......... . . .. . . . . . . . . . .. . ......... ......... . .. .. . . . . . . . . . . . d k . . . .. . .. ... ... . . . . . . . . . . . . . . . . . . . . . . •C .......... ........ . .. . .. . . . . .. . .. . ......... ......... d . . . . . . . . . . . . . . .. ..... ......... ..... 5 d 2 D Solution: First, we will determine the angular velocity of Friction Disk B. This will provide information needed to determine the velocity of Weight D. Then, we will determine the angular acceleration of the friction disk and the linear acceleration of Weight D. Velocity. If there is no slip between the friction disks, then the linear velocities of the disks at the point where they touch must be identical. Denoting the radial and circumferential coordinates of a cylindrical coordinate system whose center is at the center of Friction Disk A (see figure below), the velocity at the point where it touches Friction Disk B is ˙ va = ra θa eθ = 1 d ωa i 2 5 d ωb (−i) 4 ωa k ra θa .................................................. ................................................... .................................................. .. .................................................. ....................................... ............................................... ...................................... ................................ ......................................... .................... ........ (eθ = i at θa = 3π/2) Similarly, the velocity of Friction Disk B is ˙ vb = rb θb eθ = (eθ = −i at θb = π/2) • rb va = vb vd • θb So s nce we mus have va = vb here fo ows 2 ωb = − ωa 5 2.54. CHAPTER 2, PROBLEM 54 139 The fact the ωb is negative tells us that Friction Disk B rotates clockwise. This corresponds to the fact that θ increases in the counterclockwise direction in cylindrical coordinates. Finally, the linear velocity of the weight, vd , is equal to the velocity of Pulley C at rc = d/2 and θb = π, i.e., ˙ vd = rc θb eθ = 1 d 2 2 − ωa 5 (−k) (eθ = −k at θb = π) Therefore, the linear velocity of Weight D is vd = 1 ωa d k 5 Acceleration. In general, the acceleration vector in cylindrical coordinates for circular motion is ¨ ˙ a = −rθ2 er + rθ eθ = −rω 2 er + rα eθ Now, if there is no slip between the friction disks, the linear acceleration at the point where the disks touch must be equal. This is true because ax = dvx /dt, and vxa = vxb . The acceleration of Friction Disk A at the point where the disks touch is aa = − which simplifies to 1 12 αa d i + ωa d k 2 2 Similarly, the acceleration of Friction Disk B at the point where the disks touch is aa = ab = − which simplifies to 5 52 ab = − αb d i − ωb d k 4 4 So, since the x components of the acceleration vectors for the two disks are equal, we find 1 5 αa d = − αb d 2 4 =⇒ 2 αb = − αa 5 5 2 d ωb (k) + 4 5 d αb i 4 (ab = k, eθ = −i at θa = π/2) 1 2 d ωa (−k) + 2 1 d αa i 2 (aa = −k, eθ = i at θa = 3π /2) The linear acceleration of the weight, ad , is equal to the acceleration of Pulley C at rc = d/2 and θb = π , wherefore 1 2 ad = (eθ = −k at θb = π) d − αa (−k) 2 5 Therefore, the linear acceleration of Weight D is ad = 1 αa d k 5 ...
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