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Two particles of masses 2 kg and 3 kg are connected by a rope of total length 2 m and rotate in a horizontal circle about a common center (the fixed end). The system revolves with angular velocity 10 rad/s. In the standard arrangement where mass m1 = 2 kg is at radius r1 = 1 m and mass m2 = 3 kg is at radius r2 = 2 m from the center, find the tensions T1 (in the rope segment closer to the center) and T2 (in the outer segment).
- T1 = 200 N, T2 = 600 N
- T1 = 600 N, T2 = 200 N
- T1 = 800 N, T2 = 600 N
- T1 = 600 N, T2 = 800 N
Correct answer: T1 = 800 N, T2 = 600 N
Solution
Assuming m1=2 kg at r1=1 m from centre, m2=3 kg at r2=2 m. The rope connects: centre --- [T1 segment, length 1m] --- m1 --- [T2 segment, length 1m] --- m2. For m2: net inward force = T2 = m2*omega²*r2 = 3*100*2 = 600 N. For m1: net inward force = T1 - T2 = m1*omega²*r1 = 2*100*1 = 200 N, so T1 = 600+200 = 800 N.
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