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The position vector of the centre of mass of an asymmetric uniform bar (negligible cross-section) is, for the stepped shape shown (a top arm of length L, a drop of L, then a base arm of length 2L),

Asked in JEE Main 12th Jan 1st Shift 2019 · Centre of mass by integration

Figure: Centre of mass by integration
Answer: (1) (11)/8L ̂x+3/8L ̂y

Step-by-step solution

Model the bar as three segments: top (0,L)→(L,L) [mass L, centre (/L2,L)], drop (L,L)→(L,0) [mass L, centre (L,/L2)], base (L,0)→(3L,0) [mass 2L, centre (2L,0)]. Total length 4L.

x_cm=(L·/L2+L· L+2L· 2L)/(4L)=(11)/8L.

y_cm=(L· L+L·/L2+2L· 0)/(4L)=3/8L.

So ⃗r_cm=(11)/8L ̂x+3/8L ̂y.

Why the other options are wrong

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