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ISRO VSSC Technical Assistant Mechanical held on 25/02/2018

Option 2 : \(\frac{2}{3}l\)

ISRO VSSC Technical Assistant Mechanical held on 09/06/2019

2302

80 Questions
320 Marks
120 Mins

**Explanation:**

The** center of percussion is the point** on to a pivot extended object (rod) where a perpendicular impact will produce** no reactive shock **at the pivot.

Let us consider a rod of length L and mass m and select a pivot point at a distance p from the center of mass. We strike the rod a short blow a distance b_{i} from the center of mass as shown

from Newtons' second law

\({\rm{m}}\frac{{\rm{d}}}{{{\rm{dt}}}}{{\rm{V}}_{{\rm{centre\;of\;mass\;}}}} = {\rm{F}}\) .... (I)

or \({{\rm{I}}_{{\rm{COM}}}}\frac{{{\rm{d\omega }}}}{{{\rm{dt}}}} = {\rm{F}} \times {{\rm{b}}_{\rm{i}}}\) ....(II)

now for any point P at a distance p from the center of mass, the change of its velocity due to the rotation is given by

dV_{p} = dV_{com} – pdω

\(\frac{{d{V_p}}}{{dt}} = \frac{{d{V_{COM}}}}{{dt}} - \frac{{pdw}}{{dt}}\)

from equation (I) and (II)

\(\frac{{{\rm{d}}{{\rm{V}}_{\rm{p}}}}}{{{\rm{dt}}}} = {\rm{F}}\left( {\frac{1}{{\rm{m}}} - \frac{{{\rm{p}}{{\rm{b}}_{\rm{i}}}}}{{{{\rm{I}}_{{\rm{COM}}}}}}} \right)\)

since p is point of center of mass hence\(\frac{{{\rm{d}}{{\rm{V}}_{\rm{p}}}}}{{{\rm{dt}}}}=0\)

\({\rm{F}}\left( {\frac{1}{{\rm{m}}} - \frac{{{\rm{p}}{{\rm{b}}_{\rm{i}}}}}{{{{\rm{I}}_{{\rm{COM}}}}}}} \right)=0\)

\({b_i} = \frac{{{I_{COM}}}}{{pm}}\) ....(I)

since in question, it is given that rod is pivoted at on of their endpoint hence \(p=\frac{l}{2}\)

and \({{\rm{I}}_{{\rm{COM}}}} = \frac{1}{{12}}{\rm{m}}{{\rm{l}}^2}\)

hence after putting values of I_{COM} and p in equation (I)

\(b_i=\frac{l}{6}\)

center of percussion = p + b_{i}

center of percussion = \(\frac{l}{2}+\frac{l}{6}\)

center of percussion = \(\frac{2l}{3}\)