Question

The descending pulley has a radius $20\text{cm}$ and moment of inertia $0.2{\text{kg-m}}^2$, considering it to be a disc. The fixed pulley is light and the horizontal plane is frictionless. Find the acceleration of the block if its mass is $1.0\text{kg}$. Take $g=10{\text{m/s}}^2$.

• A. $15{\text{m/s}}^2$
• B. $10.52{\text{m/s}}^2$
• C. $20.20{\text{m/s}}^2$
• D. $5{\text{m/s}}^2$

Answer 1

The correct option is B $10.52{\text{m/s}}^2$

If the pulley descends by $x$, then $2x$ length of string will be released and for the string to remain taut, the block will move by $2x$ towards right.
Hence, acceleration of block will be double the acceleration of the descending pulley.
$\because a=\frac{dx}{dt}$
If acceleration of the block is $a$, then the acceleration of descending pulley will be $\frac{a}{2}$.


On applying Newton's second law for the block,
$T_1=m_{1}a...\left(1\right)$
On applying Newton's second law for the desending pulley,
$m_{2}g-T_1-T_2=\frac{m_{2}a}{2}...\left(2\right)$
On applying equation of torque for the descending pulley about its axis,
${\tau }_{net}=I\alpha$
$r{T}_2-r{T}_1=I\alpha$
Taking anti-clockwise sense of rotation as $+ve$,
$T_2-T_1=\frac{Ia}{2r^2}...\left(3\right)$
$[\because \alpha =\frac{a_0}{r},a_0=\frac{a}{2}]$, condition for no slipping of string over pulley.

Also, the pulley is considered to be a disc.
$\Rightarrow I=\frac{m_2{r}^2}{2}$
On adding $2\times \text{Eq.}\left(1\right))+\text{Eq}.(2)+\text{Eq.}(3)$
$2T_1+m_{2}g-T_1-T_2+T_2-T_1=2m_{1}a+\frac{m_{2}a}{2}+\frac{Ia}{2r^2}$
$\Rightarrow m_{2}g=2m_{1}a+\frac{m_{2}a}{2}+\frac{Ia}{2r^2}$
$\Rightarrow \left(\frac{2I}{r^2}\right)g=2m_{1}a+\frac{Ia}{r^2}+\frac{Ia}{2r^2}$
$\Rightarrow a=\frac{\left(\frac{2I}{r^2}\right)g}{2m_1+\left(\frac{I}{r^2}\right)+\left(\frac{I}{2r^2}\right)}$

$\Rightarrow a=\frac{\left(\frac{2\times 0.20\times 10}{(0.20{)}^2}\right)}{(2\times 1)+\left(\frac{0.20}{(0.20{)}^2}\right)+\left(\frac{0.20}{2\times (0.20{)}^2}\right)}$
$\therefore a\simeq 10.52{\text{m/s}}^2$
Acceleration of the block is $10.52{\text{m/s}}^2$