Rotational Motion Question 82
Question: A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is
Options:
A) $ \frac{wx}{d} $
B) $ \frac{wd}{x} $
C) $ \frac{w(d-x)}{x} $
D) $ \frac{w(d-x)}{d} $
Show Answer
Answer:
Correct Answer: C
Solution:
- As the weight w balances the normal reactions. So, $ w=N _1+N _2 $ (i) Now balancing torque about the COM, i.e.
anti-clockwise momentum = clockwise momentum
$ \Rightarrow $ $ N _1x=N _2(d-x) $ Putting the value of $ N _2 $ from Eq.
(i), we get $ N _1x=(w-N _1)(d-x) $
$ \Rightarrow $ $ N _1x=wd-wx-N _1d+N _1x $
$ \Rightarrow $ $ N _1d=w(d-x) $
$ \Rightarrow $ $ N _1=\frac{w(d-x)}{d} $
