Question

A 3.65 m long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]

Answer 1

Initially the pole ,starting from rest, has potential energy mg(L/2).
Just before it hits the ground all of the potential energy is changed to kinetic energy, (1/2)Iw^2
The moment of inertia of a pole about one end is (1/3)mL^2
The Energy eq. is;

KE(bottom) = PE(top)

(1/2)(1/3)mL^2w^2 = mg(L/2)

The linear velocity, v , of the end of the pole is related to the pole's angular velocity by;

v = Lw

The energy eqation becomes;

(1/3)v^2 = gL

v = SqRt(3gL)

= 10.36 m/s