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NET The uniform rod of length 4b and mass m is bent into the shape shown. The diameter od is small compared with its length.
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Given that; Maus of rod (m) = 63 kg length of rod (45)= 4x 500 mm El 6=0.5m - 2000 mm (01) 2m Therefore marr perlength and ma(gy) = (etegy), + (Tyyh @uyla )5+0+14) 5° (ayy) = mblr 6.3x (0.7) Igy = 0.2625 kami (122) = (2x2), 1T2zl2 +{I22 ), (10)6+ (09Feel free to use comment section incase you need any help.

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