Question

Please answer both questions

1) A chimney (length 24.2 m, mass 401 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

2)A chimney (length 13 m, mass 922 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 very top of the chimney, in m/s, just as it hits the ground.  

Answer 1

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=24.2 m. So, v=24.2/2*(3*9.8/24.2)1/2=13.34 m/s

2)

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.

So, velocity of top of rod=L*(3g/L)1/2

In the given problem, L=13 m. So, v=13*(3*9.8/13)1/2=19.55 m/s