For the given figure find the acceleration of | vedclass

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Question

Consider the figure,

$m_{1} g-T_{1}=m_{1} a$ $...(I)$

$T_{2}-m_{2} g=m_{2} a$ $...(II)$

$T_{1} R-T_{2} R=I \alpha$

$=\frac{M R^{2}}{2} \cd

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Consider the figure,

$m_{1} g-T_{1}=m_{1} a$ $...(I)$

$T_{2}-m_{2} g=m_{2} a$ $...(II)$

$T_{1} R-T_{2} R=I \alpha$

$=\frac{M R^{2}}{2} \cdot \frac{a}{R^{2}}$

$T_{1}-T_{2}=\frac{M a}{2}$ $...(III)$

Add equation $(I), (II),$ and $(III).$

$m_{1} g-m_{2} g+T_{2}-T_{1}=\left(m_{1} a+m_{2} aight)$

$10-5=\frac{2 a}{2}+\frac{a+a}{2}$

$5=\frac{5 a}{2}$

$a=2 
eq$

For the given figure find the acceleration of $1\, kg$ block if string is massless and mass of the pulley is $2\, kg$ and diameter of puller is $0.2\, m$ (in $m / s ^{2}$)

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