A 60-g and a 80-g ball are fixed to the ends of a 28-cm rod of negligible mass. The rod is balanced by a pivot such that it is horizontal and stays at rest. With what magnitude of angular acceleration

Question

A 60-g and a 80-g ball are fixed to the ends of a 28-cm rod of negligible mass. The rod is balanced by a pivot such that it is horizontal and stays at rest. With what magnitude of angular acceleration (in rad / s 2 ) does it start to move, if the pivot is shifted 2 cm closer to the heavier ball? (Take g = 9.8 m / s 2 )

TestHub · Accepted Answer

Balance condition 80 ℓ = 60 ( 28 − ℓ ) places the original pivot 12 cm from the 80 g ball. Shifted 2 cm closer, the arms are 10 cm and 18 cm, so τ = ( 0.06 ) ( 9.8 ) ( 0.18 ) − ( 0.08 ) ( 9.8 ) ( 0.10 ) = 0.02744 N m. I = 0.08 ( 0.10 ) 2 + 0.06 ( 0.18 ) 2 = 0.002744 kg m 2 , so α = I τ ​ = 10 rad/ s 2 .

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