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Question Number 33214 by Tinkutara last updated on 13/Apr/18

Commented by Tinkutara last updated on 13/Apr/18

Commented by Tinkutara last updated on 15/Apr/18

But if we consider the torque of mg instead of ma it still gives approximately the same answer. Which one is conceptually correct?

Commented by ajfour last updated on 14/Apr/18

Commented by ajfour last updated on 14/Apr/18

$$mg-T=ma$$$$TR=I\left(\frac{a}{R}\right)$$$$\Rightarrow mg=a[m+\frac{I}{R^2}]...\left(1\right)$$$$s=\frac{1}{2}{at}^2\Rightarrow a=\frac{2s}{t^2}$$$$usingin\left(1\right)$$$$mg=\frac{2s}{t^2}(m+\frac{I}{R^2})$$$$orI=(\frac{{mgt}^2}{2s}-m)R^2$$$$=(\frac{0.5\times 10\times 16}{2\times 1.5}-0.5){(0.15)}^2$$$$=(\frac{80}{3}-0.5)(0.0225)kg.m^2$$$$=(\frac{80}{3}-\frac{1}{2})\left(\frac{9}{400}\right)kg.m^2$$$$=\frac{157}{6}\times \frac{9}{400}=\frac{157\times 3}{800}=\frac{4.71}{8}$$$$=0.59kg.m^2.$$

Commented by ajfour last updated on 15/Apr/18

$$iftorqueofmgistakennet$$$$momentofinertiaofsystem$$$$aboutfixedaxisofpulleyis$$$$I_{system}=I+{mR}^2$$$$mgR=(I+{mR}^2)\left(\frac{a}{R}\right).$$

Commented by Tinkutara last updated on 15/Apr/18

Both are same in value approximately. Which is correct then?

Commented by ajfour last updated on 15/Apr/18

$$havesamevalueexactly.howapproximately?$$

Commented by Tinkutara last updated on 15/Apr/18

Thank you very much Sir! I got the answer. ��������

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