Consider a uniform rod of mass M and length L, that is attached to a vertical wa
ID: 1489075 • Letter: C
Question
Consider a uniform rod of mass M and length L, that is attached to a vertical wall. The rod is free to rotate in 2D from the end that is attached to the wall. Initially, the rod is horizontal and at rest, i.e. it makes and angle /2 with the vertical. The it is released. When it makes and angle with the vertical: (a) What is its angular velocity? (Hint: Make use of conservation of energy) (b) What is the linear velocity of the centre of mass of the rod? (c) What is the acceleration vector (both tangential and central) acceleration of the CM? (d) What is the force exerted on the rod by the wall at the point that the rod is attached to the wall? (Hint: The acceleration of the centre of mass is determined only by the total force acting on the rod.)
Explanation / Answer
a) height descended by centre of mass of rod.
h = L/2 cos@
using energy conservation,
loss in gravitational PE = gain in KE
mgh = Iw^2 /2
M * g * Lcos@ / 2 = ( M L^2 / 3) w^2 /2
w = sqrt (3 g cos@ / L )
b) v = w*r = sqrt(3gcos@/L) * (L/2)
v = sqrt(3gLcos@/4)
c) torque = I x alpha
mgLsin@ / 2 = (m L^2 /3 ) * alpha
alpha = 3gsin@ / 2L
Tangential acc. = alpha*r = 3gsin@ / 2L * L/2
a_t = 3gsin@ / 4
central acc. a_c = w^2 r = 3gcos@ / L * L/2 = 3gcos@ / 2