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A small trolley of mass 2.0kg resting on a horizontal turn table is connected by a light spring to the centre of the table. When the turn table is set into rotation a speed of 600rpm, the length of stretched spring is 50cm. If the original length of the spring is 42cm, find the force constant of the spring. |
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Answer» Here, mass of trolley, `m=2.0kg,` Frequency of rotation, `v=(600)/(60)=10rps` Extension produced in string, `x=50-42=8cm =8xx10^(-2)m` When the turn table is set into rotation, the tension (restoring force), in the spring is equal to the centrifugal force. Then length of stretched spring, `r=50cm=0.50m.` `:. F=kx=mr(2piv)^(2)=4pi^(2)mrv^(2)` or `k=(4pi^(2)mrv^(2))/(x)` `=(4xx(22//7)^(2)xx2.0xx0.50xx(10)^(2))/(8xx10^(-2))` `=49387.7Nm^(-1)` |
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