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A ROD STANDING VERTICALLY ON A ROUGH SURFACE BEGINGS TO FALL .FIND THE ACCELERATION OF CENTRE OF MASS JUST BEFORE TOUCHING THE GROUND.(MASS=M ,LENGTH =L )
hii rohit
Assume that the rod is rotating about the point of contact with the ground.Now ,when its other end(free end) is about to touch the ground the Centre of mass (CM) of rod will have both Tangential acceleration and Normal acceleration.So now we should calculate the above two accelrations.
For finding tangential acceleration:
Σ T =Iα. (I=inertia of rod about the contact point i.e point of rotation, α=tangential acceleration, T- net torque)
The only force causing moment is Self weight of Rod.
T=m*g*(L/2);
I= (m*L^2/12)
by substituting this in above equation you get α(alpha)=(3*g)/(2*L).
For finding Normal acceleration:
an=r*Ωf^2; (r=L/2; since CM of rod is located at L/2)
use the kinematics equation to find Ωf:
Ωf^2-Ωi^2=2*α*Θ
when its just about touch ground Θ=pi/2;
Ωi=0;
we've already calculated α(alpha)
substitute all these values in the above equation you'll get Ωf.
substitute Ωf in an=r*Ωf^2 you'll get an.
atotal= (an^2+α^2).
feel free to ask queries.
Thanks.
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