1How did you get ((v/2)T,L/2)?
The rod moves perpendicular to the x-axis and also rotates about the centre of mass.
A light rigid rod is placed on a smooth horizontal surface. Initially the end A begins to move vertically upward with constant velocity v and centre of the rod upward with a velocity v/2 having downward acceleration a/2. The other end moves downward with
a) 0 initial velocity having zero acceleration
b) 0 initial velocity having a downward acceleration
c) non-zero initial velocity and zero downward acceleration
d) all of the above.
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4 Answers
1i think (B)
take the rod along x axis,with centre of rod at the origin.consider at any time T ,the cordinate of one end of the rod,and centre of the rod as a ((v/2)T,L/2) and (1/2(a/2)T2,0)
take other rod end have velocity X,and acceleration K.
find its cordinate as a function of T.(XT+1/2 KT2,-L/2)
Now,use the fact the the rod is COLINEAR..i.e: area of traingle is 0.
11answer sahi hai kya?
okay ,lets consider those as three points(not a line/rod).For them to be always colinear,the same condition should be satisfied.