- #1
wolf party
- 24
- 0
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 theorm 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
i used parallel axis theorm 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
Last edited: