Question

The descending pulley has a radius $20\mathrm{cm}$ and moment of inertia $0.20\mathrm{kg}-{\mathrm{m}}^2$. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is $1.0\mathrm{Kg}$.

Answer 1

Step 1. Given data:

It is given that, mass is $1.0\mathrm{Kg}$, radius($r$) is $20\mathrm{cm}$, and the moment of inertial is $0.20\mathrm{kg}-{\mathrm{m}}^2$.

Tension $=$ $\mathrm{T}$ (Plane) and $\mathrm{T}\text{'}$ (side)

Step 2. Formula to be used

From the below figure, the block is attached to $1$ string. The pulley is attached to $2$ strings. There is weight on the pulley. As a result, it will accelerate both linearly and angularly. Using the torque equation, we can determine that the tensions in the $2$ strings holding the $2$ pulleys together are different.

The formula of torque is,

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

Here, $\mathrm{\tau }$ is torque, $\mathrm{I}$ is the moment of inertia, and $\mathrm{\alpha }$ is the angular acceleration.

Step 3. Determine the acceleration of the pulley.

From the above figure, $2$ strings are attached to the pulley and $1$ string is attached to the block, their acceleration will vary but is dependent on one another.

The block travels $\text{'}2\mathrm{x}\text{'}$ distance right if the pulley moves $\text{'}\mathrm{x}\text{'}$ distance down. Therefore, if the block accelerates by $\mathrm{a}$, the pulley will accelerate by $\frac{\mathrm{a}}{2}$.

So, the tension of the string connecting block is ${\mathrm{T}}_1$.

Let us consider that, the mass of the block is ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ is the mass of pulley.

Acceleration of the massive pulley will be half of that of the block.

So,

${\mathrm{T}}_1={\mathrm{m}}_1\mathrm{\alpha }$

$=\left(1\right)\mathrm{\alpha }$

$=\mathrm{\alpha }$

We have,

${\mathrm{m}}_2=\mathrm{r\alpha }$

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

So,

$\mathrm{\tau }=\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\mathrm{r}$ ----------- $\left(2\right)$

From the given ,

$\mathrm{r}=20\mathrm{cm}$

$=0.2\mathrm{m}$

Substitute the value in equation $\left(2\right)$. We get,

$\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\mathrm{r}=\mathrm{I\alpha }$

$\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\left(0.2\right)=\left(0.2\right)\mathrm{\alpha }$

$\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=\mathrm{\alpha }$

Now,

${\mathrm{m}}_2=\mathrm{r\alpha }$

$\frac{\mathrm{\alpha }}{2}=0.2\mathrm{\alpha }$

$\mathrm{\alpha }=2.5\mathrm{\alpha }$

We know that,

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

$2.5\mathrm{\alpha }=\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)$

Therefore,

$\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-2.5\mathrm{\alpha }$

Step 4. Newton's second law for a pulley.

We have got angular acceleration in terms of $\mathrm{\alpha }$ and difference in tensions in terms of $\mathrm{\alpha }$.

We will use Newton's second law for a pulley.

$\Rightarrow$ $\mathrm{I}=\frac{{\mathrm{mr}}^2}{2}$

$\Rightarrow$ $\mathrm{m}=\frac{2\times 0.2}{{\left(0.2\right)}^2}$

$\Rightarrow$ $\mathrm{m}=10\mathrm{kg}$

$\Rightarrow$${\mathrm{m}}_2\mathrm{g}-\left({\mathrm{T}}_1+{\mathrm{T}}_2\right)=\mathrm{m}\times {\mathrm{m}}_2$

$\Rightarrow$ $=\mathrm{M}\frac{\mathrm{\alpha }}{2}$

$\Rightarrow$ ${\mathrm{T}}_1-{\mathrm{T}}_2=-2.5\mathrm{\alpha }$

$\Rightarrow$ ${\mathrm{T}}_1=\mathrm{\alpha }$

$\Rightarrow$ $9.5\mathrm{\alpha }={\mathrm{m}}_2\mathrm{g}$

On replacing the value of ${\mathrm{m}}_2$ using $\frac{1}{2}{\mathrm{mr}}^2=\mathrm{I}$.

So,

$9.5\mathrm{\alpha }=\mathrm{m}\mathrm{g}$

Put the value of $\mathrm{m}$ and gravitational constant $\mathrm{g}$ is $9.8\mathrm{m}/{\mathrm{s}}^2$, we get,

$\mathrm{\alpha }=\frac{10\times 9.8}{9.5}$

$\mathrm{\alpha }=10.315\mathrm{m}/{\mathrm{s}}^2$

Therefore, the acceleration of the pulley will be $10.315\mathrm{m}/{\mathrm{s}}^2$.