A) \[{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}g\]
B) \[\frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\]
C) \[\frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}-{{m}_{2}}}g\]
D) zero
Correct Answer: A
Solution :
The free body diagram showing the various forces acting on the pulley mass are as follows: Equating the vertical forces, we have \[{{m}_{1}}g-T={{m}_{1}}a\] ...(i) \[T-{{m}_{2}}g={{m}_{a}}a\] ...(ii) From Eqs. (i) and (ii), we get \[a=\frac{{{m}_{1}}g-{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}\] ...(iii) The acceleration of centre of mass is \[{{a}_{CM}}=\frac{{{m}_{1}}a-{{m}_{2}}a}{{{m}_{1}}+{{m}_{2}}}\] Putting the value of a from Eq. (iii), we get \[{{a}_{CM}}=\frac{{{({{m}_{1}}-{{m}_{2}})}^{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}g\]You need to login to perform this action.
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