Question

Consider the system shown in figure with $m_{1}=20.0 \,kg$, $m_{2}=12.5 \,kg , R=0.200 \,m$, and the mass of the pulley $M=10.0 \, kg$. Object $m_{2}$ is resting on the floor, and object $m_{1}$ is $4.00 \,m$ above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley. The time interval (in s) required for $m_{1}$ to hit the floor is ___.

Answer 1

Let $T_{1}$ represent the tension in the cord above $m_{1}$ and $T_{2}$ the tension in the cord above the lighter mass. The two blocks move with the same acceleration because the cord does not stretch, and the angular acceleration of the pulley is $a / R$. For the heavier mass we have

or $ T_{1}-m_{1} g=m_{1} a\,\,\,...(i)$
For the lighter mass,
$T_{2}-m_{2} g=m_{2} a\,\,\,...(ii)$
We assume the pulley is a uniform disk: $I=(1 / 2) M R^{2}$
$\sum \tau=I \alpha \rightarrow+T_{1} R-T_{2} R=\frac{1}{2} M R^{2}(a / R)$
or $ T_{1}-T_{2}=\frac{1}{2} M a\,\,\,...(iii)$

From (i), (ii) and (iii); $a=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}+\frac{1}{2} M}\right) g$
$=\left(\frac{20.0\, kg -12.5\, kg }{20.0\, kg +12.5 \,kg +\frac{1}{2}(10.0\, kg )}\right)\left(10.0\, m / s ^{2}\right)=2.0 \,m / s ^{2}$
Next, $x=0+0+\frac{1}{2} a t^{2} $
$\Rightarrow t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{2(4.00 m )}{2.0\, m / s ^{2}}}=2.0\, s$