Question

Two blocks of mass m each are connected by a string passing over a pulley as shown in the figure. The table is smooth. Moment of inertia of pulley is $1$. Find out acceleration of the hanging mass.

• A. $a=\frac{mg}{2m+\frac{I}{r^2}}$
  
• B. $a=\frac{mg}{m+\frac{I}{r^2}}$
• C. $a=\frac{2mg}{2m+\frac{I}{r^2}}$
• D. $a=\frac{2mg}{m+\frac{I}{r^2}}$

Answer 1

The correct option is A $a=\frac{mg}{2m+\frac{I}{r^2}}$

consider blocks to be named as m and m

acceleration "a" and tensions "t" and "t'"

for first block,

$mg-t=ma$

$t=m(g-a)$

for second block,

$t^{\prime }=ma$

net torque,

$=tr-t^{\prime }r$

$=(t-t^{\prime })r$

if angular acceleration, $\alpha$

$a=\alpha r$

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

torque

$t=I\alpha$

$(t-t^{\prime })r=I\frac{a}{r}$

$m(g-a)-ma=I\frac{a}{r^2}$

$mg=ma+ma+I\frac{a}{r^2}$

$\therefore a=\frac{mg}{2m+\frac{I}{r^2}}$