In Fig.$\mathbf{10}\mathbf{-}\mathbf{41}$ , block $\mathbf{1}$has mass ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{460}\mathbf{\text{\hspace{0.17em}g}}$, block $\mathbf{\text{2}}$has mass${\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{500}\mathbf{\text{\hspace{0.17em}g}}$ , and the pulley, which is mounted on a horizontal axle with negligible friction, has radius $\mathbf{R}\mathbf{=}\mathbf{5}\mathbf{.00}\mathbf{\text{\hspace{0.17em}cm}}$. When released from rest, block $\mathbf{\text{2}}$ falls$\mathbf{75}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$ in$\mathbf{5.00}\mathbf{\text{\hspace{0.17em}s}}$ without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension${\mathbf{\text{T}}}_{\mathbf{2}}$ and (c) tension${\mathbf{\text{T}}}_{\mathbf{1}}$ ? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia?

1. The mass of block 1 is,${\mathrm{m}}_1=460\text{\hspace{0.17em}g}=0.460\text{\hspace{0.17em}kg}$.
2. The mass of block 2 is,${\mathrm{m}}_2=500\text{\hspace{0.17em}g}=0.500\text{\hspace{0.17em}kg}$.
3. The distance travelled by block vertically by y axis is,$\mathrm{y}=75\text{\hspace{0.17em}cm}=0.750\text{\hspace{0.17em}m}$.
4. The time is,$\mathrm{t}=5\text{\hspace{0.17em}sec}$.
5. The gravitational acceleration is,$\mathrm{g}=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.
6. The radius is, $\mathrm{R}=5\text{\hspace{0.17em}cm}=0.05\text{\hspace{0.17em}m}$.

Using kinematic equations and given height and time, you can find the acceleration of the masses. Using Newton’s second law of motion, you can find the tension in the string. And using the tensions, it is possible to find the net torque and acceleration of the pulley. The formulas used are given below.

1. Kinematical equation, $\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$
2. Newton’s second law of motion, $\mathbf{T}\mathbf{=}{\mathbf{m}}_{\mathbf{2}}\mathbf{g}\mathbf{-}{\mathbf{m}}_{\mathbf{2}}\mathbf{a}$
3. Tangential acceleration, $\mathbf{a}\mathbf{=}\mathbf{\alpha }\mathbf{\times }\mathbf{R}$
4. Net torque,$\mathbf{\tau }\mathbf{=}\mathbf{(}{\mathbf{T}}_{\mathbf{2}}\mathbf{-}{\mathbf{T}}_{\mathbf{1}}\mathbf{)}\mathbf{R}$
5. Torque,$\mathbf{\tau }\mathbf{=}\mathbf{I\alpha }$

To calculate the acceleration of block 2, let’s assume that it moves along y-axis and ‘a’ be the acceleration ofthe heavier block ${\mathrm{m}}_2$in the downward direction. Assuming downward direction as positive,from the kinematics equation

$\begin{array}{c}\mathrm{y}=\frac{1}{2}{\mathrm{at}}^{\text{2}}\\ \mathrm{a}=\frac{2\mathrm{y}}{{\mathrm{t}}^2}\end{array}$

Substitute all the value in the above equation

$\begin{array}{c}\mathrm{a}=\frac{2\left(0.750\text{\hspace{0.17em}m}\right)}{{\left(5\text{\hspace{0.17em}s}\right)}^2}\\ =6\times {10}^{-2}\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

Hence the acceleration of block is, $6\times {10}^{-2}\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.

To find tension on block 2, we will consider Newton’s second law of motion,

$\begin{array}{c}\mathrm{T}={\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_2\mathrm{a}\\ \mathrm{T}={\mathrm{m}}_2(\mathrm{g}-\mathrm{a})\end{array}$

Substitute all the value in the above equation

$\begin{array}{c}{\mathrm{T}}_2=\left(0.500\text{\hspace{0.17em}kg}\right)(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}-6\times {10}^{-2}\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})\\ =4.87\text{N}\end{array}$

Hence the tension is, $4.87\text{N}$.

Tofind tension on block 1, we will again consider Newton’s second law of motion,

$\begin{array}{c}{\mathrm{m}}_1\mathrm{g}-{\mathrm{T}}_1=-{\mathrm{m}}_1\mathrm{a}\\ {\mathrm{T}}_1={\mathrm{m}}_1(\mathrm{g}+\mathrm{a})\end{array}$

Substitute all the value in the above equation

$\begin{array}{c}{\mathrm{T}}_1=\left(0.460\text{\hspace{0.17em}kg}\right)(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}+6\times {10}^{-2}\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})\\ =4.54\text{N}\end{array}$

Hence the tension is, $4.54\text{N}$.

Tofind the tangential acceleration of point on rim of the pulley we will consider the acceleration of blocks also. They might give similar values,

$\begin{array}{c}\mathrm{a}=\mathrm{\alpha }\times \mathrm{R}\\ \mathrm{\alpha }=\frac{\mathrm{a}}{\mathrm{R}}\end{array}$

Substitute all the value in the above equation

$\begin{array}{c}\mathrm{\alpha }=\frac{6\times {10}^{-2}\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{5\times {10}^{-2}\text{\hspace{0.17em}m}}\\ =1.20\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

Hence the angular acceleration is, $1.20\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$.

To find the rotational inertia of the pulley, we have the formula for torque,

$\mathrm{\tau }=\mathrm{I\alpha }$

Net torque,

$\mathrm{\tau }=({\mathrm{T}}_2-{\mathrm{T}}_1)\mathrm{R}$

Comparing both the formulas we get,

$\begin{array}{c}({\mathrm{T}}_2-{\mathrm{T}}_1)\mathrm{R}=\mathrm{I\alpha }\\ \mathrm{I}=\frac{({\mathrm{T}}_2-{\mathrm{T}}_1)\mathrm{R}}{\mathrm{\alpha }}\end{array}$

Substitute all the value in the above equation

$\begin{array}{c}\mathrm{I}=\frac{(4.87\text{\hspace{0.17em}N}-4.54\text{\hspace{0.17em}N})\times 5\times {10}^{-2}\text{\hspace{0.17em}m}}{1.20\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}}\\ =1.38\times {10}^{-2}{\text{\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

Hence the rotational inertia is, $1.38\times {10}^{-2}{\text{\hspace{0.17em}kg.m}}^{\text{2}}$.