(a) If $\mathbf{R}\mathbf{=}\mathbf{}\mathbf{12}\mathbf{\text{\hspace{0.17em}cm}}$ ,$\mathbf{M}\mathbf{=}\mathbf{}\mathbf{400}\mathbf{\text{\hspace{0.17em}g}}$ , and$\mathit{m}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{\text{\hspace{0.17em}g}}\mathbf{}$ in Fig.$\mathbf{10}\mathbf{-}\mathbf{19}$ , find the speed of the block after it has descended$\mathbf{\text{50\hspace{0.17em}cm}}$ starting from rest. Solve the problem using energy conservation principles.

(b) Repeat (a) with $\mathbf{}\mathit{R}\mathbf{=}\mathbf{}\mathbf{5.0}\mathbf{\text{\hspace{0.17em}cm}}$.

$\mathrm{R}=12\text{\hspace{0.17em}cm}=0.12\text{\hspace{0.17em}m}$

$\mathrm{M}=400\text{\hspace{0.17em}g}=0.4\text{\hspace{0.17em}kg}$

$\mathrm{m}=50\text{\hspace{0.17em}g}=0.05\text{\hspace{0.17em}kg}$

$\mathrm{y}=50\text{\hspace{0.17em}cm}=0.5\text{\hspace{0.17em}m}$

You can write the expression for tension by applying Newton’s second law to the given system and torque on the system. Equating them, you will get the acceleration of the block. Then using the law of conservation of energy, you will get an expression for speed. Inserting values in it, you can get thespeed of the block if$\mathbf{R}\mathbf{=}\mathbf{12}\mathbf{}\mathbf{\text{\hspace{0.17em}cm}}\mathbf{}$ and if$\mathbf{R}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{\text{\hspace{0.17em}cm}}\mathbf{}$ All the formulas used are given below

Applying Newton’s second law to the given system, we get

$\begin{array}{c}{\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}\\ \mathrm{T}-\mathrm{mg}=\mathrm{ma}\end{array}$

$\mathrm{T}=\mathrm{mg}+\mathrm{ma}$ …(1)

The torque on the pulley is

$\mathrm{\tau }=-\mathrm{TR}$

(Negative sign indicates thatthepulley is rotating in a clockwise direction from rest.)

Also,

$\begin{array}{c}\mathrm{\tau }=\mathrm{I\alpha }\\ -\mathrm{TR}=\frac{1}{2}{\mathrm{MR}}^2\mathrm{\alpha }\end{array}$

But,

$\mathrm{\alpha }=\frac{\mathrm{a}}{\mathrm{r}}$

$\mathrm{T}=-\frac{1}{2}\mathrm{Ma}$ …(2)

Equating (1) and (2), we get

$\begin{array}{c}\mathrm{mg}+\mathrm{ma}=-\frac{1}{2}\mathrm{Ma}\\ \mathrm{ma}+\frac{1}{2}\mathrm{Ma}=-\mathrm{mg}\\ \left(\mathrm{m}+\frac{1}{2}\mathrm{M}\right)\mathrm{a}=-\mathrm{mg}\\ \mathrm{a}=-\frac{\mathrm{mg}}{\left(\mathrm{m}+\frac{1}{2}\mathrm{M}\right)}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{a}=-\frac{\left(0.05\text{\hspace{0.17em}kg}\right)(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})}{0.05\text{\hspace{0.17em}kg}+\left(0.5\right)\times 0.4\text{\hspace{0.17em}kg}}\\ =-1.96\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

Magnitude of the acceleration is, $\mathrm{a}=1.96\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$

According to the conservation of energy,

${\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$

In this case,

$\begin{array}{c}\text{may}=\frac{1}{2}{\text{mv}}^2\\ \text{ay}=\frac{1}{2}{\text{v}}^2\\ \text{v}=\sqrt{2\text{ay}}\\ =\sqrt{2(1.96\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})\left(0.5\text{\hspace{0.17em}m}\right)}\\ =1.4\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore, the speed of the block if $\mathrm{R}=12\text{\hspace{0.17em}cm}$ is,$1.4\text{\hspace{0.17em}m}/\text{s}$ .

Sincethespeed of the block is independent of R, the speed will bethesame if $\mathrm{R}=5\text{cm}$.

Therefore, the speed of the block is, $1.4\text{\hspace{0.17em}m}/\text{s}$.