A) \[\frac{m}{M}L\,\cos \,\theta \]
B) \[\frac{m}{M+m}L\]
C) \[\frac{M+m}{mL\,\cos \,\theta }\]
D) \[\frac{mL\,\cos \,\theta }{m+M}\]
Correct Answer: D
Solution :
Here, the x-coordinate of centre of mass of the system remains unchanged when the mass m moved a distance \[L\cos \theta \], let the mass (m + M) moves a distance x in the backward direction. (from the theory of centre of mass, change in coordinate of centre of mass = 0) \[\therefore \] \[(M+m)\,x-mL\,\,\cos \,\theta =0\] \[\therefore \] \[x=\frac{mL\,\cos \,\,\theta }{m+M}\]You need to login to perform this action.
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