First slide

Law of conservation of angular momentum

Question

End B of a thin rod of length ‘l’ is hinged so that it can swing in a vertical plane. ‘S’ is a stopper at a distance ‘x’ below the hinge B. Initially the rod is held in horizontal position by an external agent as shown in figure. After the rod is released it swings in a vertical plane and when it becomes vertical it strikes the stopper S and stops. If during the course of impact the horizontal component of reaction at the hinge B is zero, find x.

Difficult

A

$\frac{l}{2}$
B

$\frac{2 l}{3}$
C

$\frac{3 l}{4}$
D

$\frac{5 l}{6}$

Solution

During the course of impact moment of all external forces about point S is zero and hence angular momentum of the rod about S is conserved.
Just after the impact angular momentum of the rod about S, L_f = 0.
Linear momentum of the element of length dr is given by,
$dP = dm \times V = (\frac{m}{l} dr) (ωr) = \frac{mω}{l} rdr$
Therefore angular momentum of the rod about S before impact is given by
$L_{i} = \int_{0}^{x} (x - r) dP - \int_{x}^{l} (r - x) dP$
$= x \int_{0}^{x} (\frac{mω}{l}) rdr - \int_{0}^{x} (\frac{mω}{l}) r^{2} dr - \int_{x}^{l} (\frac{mω}{l}) r^{2} dr + x \int_{x}^{l} (\frac{mω}{l}) rdr$
$= (\frac{mω}{l}) [\frac{x^{3}}{2} - \frac{x^{3}}{3} - \frac{l^{3}}{3} + \frac{x^{3}}{3} + \frac{{xl}^{2}}{2} - \frac{x^{3}}{2}]$
$= \frac{mω}{l} [\frac{{xl}^{2}}{2} - \frac{l^{3}}{3}]$
Now $L_{i} = L_{f} \Rightarrow \frac{mω}{l} (\frac{{xl}^{2}}{2} - \frac{l^{3}}{3}) = 0 \Rightarrow x = \frac{2 l}{3}$

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