A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.40). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks?

It is given that the mass of the block as $m=0.130\mathrm{kg}$, initial radius as $r_i=0.800\mathrm{m}$, initial tangential speed as $v_i=4.00\mathrm{m}/\mathrm{s}$ and the breaking force as $T_{break}=30.0\mathrm{N}$.

The centripetal force is given by the tension in the string as $T=\frac{m{v}^2}{r}$ and the conservation of angular momentum in any given time is $L=mvr$.

Thus, the centripetal force can be written as follows:

$$\begin{gathered} T=\frac{1}{m{r}^2}\frac{{\left(mvr\right)}^2}{r} \\ =\frac{{\left(mvr\right)}^2}{m{r}^3} \\ =\frac{L^2}{m{r}^3} \end{gathered}$$

From the centripetal force, the radius of the circle is,

$$\begin{gathered} r^3=\frac{L^2}{mT} \\ r={\left(\frac{L^2}{mT}\right)}^{\frac{1}{3}} \end{gathered}$$

Since, $L=m{v}_i{r}_i$ then the radius of the circle just before the string breaks is .

role="math" localid="1667990349167" $r_{break}={\left(\frac{m{N}_i^2{r}_j^2}{T_{break}}\right)}^{\frac{1}{3}}$

Substitute all the known values in the above equation and simplify.

$$\begin{gathered} r_{break}={\left(\frac{m{N}_i^2{r}_j^2}{T_{break}}\right)}^{\frac{1}{3}} \\ ={\left(\frac{(0.130\mathrm{kg})(4.00\mathrm{m}/\mathrm{s}{)}^2(0.800\mathrm{m}{)}^2}{30.0\mathrm{N}}\right)}^{\frac{1}{3}} \\ ={\left(\frac{\left(0.130\mathrm{kg}\right)(16.00\mathrm{m}/\mathrm{s})\left(6.400\mathrm{m}\right)}{30.0\mathrm{N}}\right)}^{\frac{1}{3}} \\ =0.354\mathrm{m} \end{gathered}$$

Therefore, the radius of the circle when the string breaks is 0.354m.