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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 at a speed of 360 rpm,, the length of the stretched spring is 43cm. If the original length of the spring is 36cm, determing the force constant of the spring. |
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Answer» Here, `m=2.0kg, v=(360)/(60)=6rps` Extension produced in spring , `y=43-36=7cm=7xx10^(-2)m` When turn table is set into rotation, the tension in the spring will provide the required centripetal force i.e., `ky=mr(2piv)^(2)=4pi^(2)v^(2)mr` [where r is the length of stretched spring`=43cm`] or `k=(4pi^(2)v^(2)mr)/(y)` `=(4xx(22//7)^(2)xx6^(2)xx(2.0)xx(43xx10^(-2)))/((7xx10^(-2))` `=17475Nm^(-1)` |
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