Deriving the Moment of Inertia for a Rod with a Pivot at One End

[wolf party]

a plank is conected horizontally at both ends to two points. One of the points dissapears and the other turns into a pivot. show that the Planck of length l,mass m pivoted at one end has an acceleration (3/2)*g initially



i used parallel axis theorem do derive the moment of inertia of a rod with pivot at one end
I = 1/3 (M*L*L)

the torque acting at the CM of rod
TAU=-Mg(postion of cm)=-Mg(x(cm))=-(0.5*l)Mg

Newtons second law in rotation

ALPHA = TAU/I = (-0.5*l)Mg/(1/3*M*L*L) = (-) 3/2*g*l


this is not what the question wanted as i have an extra l, what have i missed? thanks

 

[wolf party]

anybody?

 

[Mr.V]

Your answer is correct, but they are asking for the acceleration which is equal to ALPHA*l

 

[wolf party]

cheers