A block with mass m is revolving with linear speed ${\mathit{v}}_{\mathbf{1}}$ in a circle of radius${\mathit{r}}_{\mathbf{1}}$ on a frictionless horizontal surface (see Fig. E10.40). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to ${\mathit{r}}_{\mathbf{2}}$. (a) Calculate the tension T in the string as a function of r, the distance of the block from the hole. Your answer will be in terms of the initial velocity ${\mathit{v}}_{\mathbf{1}}$and the radius ${\mathit{r}}_{\mathbf{1}}$. (b) Use $\mathit{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{r}}_{\mathbf{1}}}^{{\mathbf{r}}_{\mathbf{2}}}\overrightarrow{\mathbf{T}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{.}\mathit{d}\overrightarrow{\mathbf{r}}$to calculate the work done by $\overrightarrow{\mathbf{T}}$when r changes from ${\mathit{r}}_{\mathbf{1}}$to ${\mathit{r}}_{\mathbf{2}}$. (c) Compare the results of part (b) to the change in the kinetic energy of the block.

It is given that the mass of block as m, initial linear speed as $v_1$ , initial radius as$r_1$ and final radius as$r_2$ .

The tension force provide a centripetal force then $T=\frac{m{v}^2}{r}$and the angular momentum of the force is $L=m{v}_1{r}_1$.

Find speed from the angular momentum as follows:

$$\begin{gathered} mvr=m{v}_1{r}_1 \\ \Rightarrow v=\frac{v_1{r}_1}{r} \end{gathered}$$

Substitute the value of v in T and simplify.

$\begin{array}{r}T=\frac{m{\left(\frac{v_1{r}_1}{r}\right)}^2}{r}\\ =\frac{m{v}_1^2{r}_1^2}{r^3}\end{array}$

Therefore, the required tension force is $T=\frac{m{v}_1^2{r}_1^2}{r^3}$.

It is given that the work done by the tension force as $W={\int }_{r_1}^{r_2} \overrightarrow{T}\left(r\right)\cdot d\overrightarrow{r}$.

Since, $\overrightarrow{T}\left(r\right)$ and $d\overrightarrow{r}$are always opposite in direction implies $\overrightarrow{T}\left(r\right).d\overrightarrow{r}=-T\left(r\right)dr$. Thus the work done can be written as follows:

$\begin{array}{r}W=-{\int }_{r_1}^{r_2} T\left(r\right)\cdot dr\\ ={\int }_{r_1}^{r_2} \frac{m{v}_1^2{r}_1^2}{r^3}dr\\ =m{v}_1^2{r}_1^2{\int }_{r_1}^{r_2} \frac{1}{r^3}dr\\ =m{v}_1^2{r}_1^2{\left(\frac{-2}{r^2}\right)}_{r_1}^{r_2}\end{array}$

Further, simplify as follows:

$\begin{array}{r}W=-m{v}_1^2{r}_1^2{\left(\frac{-1}{2r^2}\right)}_{r_1}^{r_2}\\ =m{v}_1^2{r}_1^2{\left(\frac{1}{2r^2}\right)}_{r_1}^{r_2}\\ =m{v}_1^2{r}_1^2\left(\frac{1}{2r_2^2}-\frac{1}{2r_1^2}\right)\\ =\frac{m{v}_1^2{r}_1^2}{2}\left(\frac{1}{r_2^2}-\frac{1}{r_1^2}\right)\end{array}$

Therefore, the work done by the tension force is $W=\frac{m{v}_1^2{r}_1^2}{2}\left(\frac{1}{r_2^2}-\frac{1}{r_1^2}\right)$.

The change in kinetic energy is given by $∆K=K_2-K_1$. Here,$K_1=\frac{m{v}_1^2}{2}$and$K_2=\frac{m{v}_2^2}{2}$then,$∆K=\frac{m{v}_2^2}{2}-\frac{m{v}_1^2}{2}$

Find speed from the angular momentum as follows:

$\begin{array}{r}m{v}_1{r}_1=m{v}_2{r}_2\\ \Rightarrow v_2=\frac{v_1{r}_1}{r_2}\end{array}$.

Substitute $v_2=\frac{v_1{r}_1}{r_2}$in $∆K$and simplify.

$\begin{array}{r}\mathrm{\Delta }K=\frac{m{\left(\frac{v_1{r}_1}{r_2}\right)}^2}{2}-\frac{m{v}_1^2}{2}\\ =\frac{m}{2}\left(\frac{v_1^2{r}_1^2}{r_2^2}-v_1^2\right)\\ =\frac{m{v}_1^2}{2}\left(\frac{r_1^2}{r_2^2}-1\right)\\ =\frac{m{v}_1^2{r}_1^2}{2}\left(\frac{1}{r_2^2}-\frac{1}{r_1^2}\right)\end{array}$

Therefore, the change in kinetic energy is equal to work done that is $∆K=W$.