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} $
