Kinematics Question 732

Question: Two masses $ M $ and $ m $ are attached to a vertical axis by weightless threads of combined length $ l $ . They are set in rotational motion in a horizontal plane about this axis with constant angular velocity $ \omega $ . If the tensions in the threads are the same during motion, the distance of $ M $ from the axis it’s [MP PET 1995]

Options:

A) $ \frac{Ml}{M+m} $

B)$ \frac{ml}{M+m} $

C) $ \frac{M+m}{M}l $

D)$ \frac{M+m}{m}l $

Show Answer

Answer:

Correct Answer: B

Solution:

If the both mass are revolving about the axis yy’ and tension in both the threads are equal then

$ M{{\omega }^{2}}x=m{{\omega }^{2}}(l-x) $

therefore $ Mx=m(l-x) $

therefore $ x=\frac{ml}{M+m} $



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