Question

A chimney (length 30 m, mass 962 kg] cracks at the base and topples.  Assume:

- the chimney behaves like a thin rod, and does not break apart as it falls.

- only gravity (no friction) acts on the chimney as it falls.

- the bottom of the chimney pivots, but does not move.

Find the linear speed of the center of mass of the chimney, in m/s, just as it hits the ground

Answer 1

Gravitational potential energy of an extended body is given by: mgh, where m is mass, g is gravitational acceleration, h is height of center of mass of the extended body.

For a uniform rod, center of mass is at the midpoint. So, assuming length of rod to be L, gravitational potential energy=mgL/2

Also, kinetic energy of a rotating object is given by: 1/2*I*2, where I is moment of inertia and is angular velocity.

For a uniform rod, potential energy about its end is given by: mL2/3, where m is mass, L is length of rod. So, kinetic energy= 1/2*(mL2/3)*2

Using conservation of energy, initial potential enrgy=final kinetic energy => mgL/2=1/2*(mL2/3)*2

=>gL=L2/3 * 2

=>2=3g/L

=>=(3g/L)1/2.

Now,the center of mass moves in a circle. In circular motion, v=r, where v is linear velocity, r is radius of circular path and is angular velocity.

The center of mass moves in a circle with radius L/2. So, velocity of center of mass=L/2*(3g/L)1/2

In the given problem, L=30 m. So, v=30/2*(3*9.8/30)1/2=14.85 m/s