A) \[\frac{3g}{2}\]
B) \[\frac{2g}{3}\]
C) \[\frac{g}{2}\]
D) \[\frac{g}{3}\]
Correct Answer: A
Solution :
\[\tau =\]force \[\times \]shortest distance\[=I\alpha \] \[mg\frac{\ell }{2}=\frac{m{{\ell }^{2}}}{3}\alpha \] \[\alpha =\frac{3g}{2\ell }\] \[{{a}_{T}}=\alpha \,\,r=\frac{3g}{2\ell }\times \ell =\frac{3g}{2}\] \[{{a}_{c}}={{w}^{2}}r=0\] \[\therefore \overrightarrow{a}=\overrightarrow{{{a}_{T}}}+\overrightarrow{{{a}_{c}}}\] \[\therefore a=\frac{3g}{2}\]You need to login to perform this action.
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