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​

Answer 1

Step 1. Given data:

Initial angular velocity = ${\omega }_0$

Initial tension in the string = $T_0$

Final length $\left(L\right)$= $2\times$Initial length$\left(L_0\right)$

Final angular velocity $\left({\omega }_{}\right)$= $2\times$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

$$\begin{gathered} T=m{\left(2{\omega }_0\right)}^2.\left(2L_0\right)=8m{{\omega }_0}^2.L_0 \\ T=8T_0 \end{gathered}$$

Thus, the final tension will become eight times the initial tension.