Question

# A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates fixed center at an angular velocity ${\omega }_{0}$ . If the length of the string and angular velocity both are doubled, the tension in the string which was initially ${T}_{0}$ is now​

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Solution

## Step 1. Given data:Initial angular velocity = ${\omega }_{0}$Initial tension in the string = ${T}_{0}$ Final length $\left(L\right)$= $2×$Initial length$\left({L}_{0}\right)$Final angular velocity $\left({\omega }_{}\right)$= $2×$initial angular velocity$\left({\omega }_{0}\right)$Let final tension in the string = $T$Step 2. Formula used:$T=m{\omega }^{2}L$Where $m$= mass.Step 3. Calculations:According to the question, putting the given values, we get$T=m{\left(2{\omega }_{0}\right)}^{2}.\left(2{L}_{0}\right)=8m{{\omega }_{0}}^{2}.{L}_{0}\phantom{\rule{0ex}{0ex}}T=8{T}_{0}$Thus, the final tension will become eight times the initial tension.

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