Given: 242=N, 183=N, 304=N, 366=N, 1507=ωRPM CW, 2502=ωRPM CCW Find: 5ωSolution: We first need to find 7N. Since 674322rrrrr+=++(1) and since “all gears have the same module” we can replace rwith Nin equation (1) and obtain 674322NNNNN+=++, (2) i.e. ( )()54363018224264327=−++=−++=NNNNN. (3) We assume that arm 5rotates CW and obtain the velocity profile for the gear system shown below (separate explanation for the velocities in gear 3 is given later). We therefore have ()3223253rrrrωωω++=, (4) ()()()()()()42276325476522325325476533325422rrrrrrrrrrrrrrrrrrrrωωωωωωωωωω+−−+=+−++++=+−++=(5) by virtue of (4), and ()6765776rrrr+−=ωωω. (6) 234677ω2ω22rω()325rr+ω55()33325rrrωω++77rω()765rr+ω()765rr+ω
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