Question

A chimney (length 10.8 m, mass 588 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

A chimney (length 13 m, mass 479 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.  

A 13.7 kg cart travels at constant speed, and its four wheels are rolling without slipping. The wheels are disks (radius 0.134 m, mass 6.05 kg), and they turn at constant ω = 26.4 rad/s. Find the total kinetic energy of the cart.

Answer 1

Multiple questions posted, so answering only the first question:

1. Given: Length of chimney=10.8 m, mass of chimney=588 kg.

Now, initially, height of center of mass of chimney=height of chimney/2. This is because the chimney is assumed to be rod and center of mass of uniform rod lies at its midpoint.

So,initial height of center of mass of chimney=10.8/2=5.4 m.

Also,finally, when the chimney falls on ground, height of center of mass of chimney becomes zero.

Now,gravitational potential energy of a rigid body is given by:mghcm, where m is mass, g is gravitational acceleration,hcm is height of center of mass.

So,initial potential energy=mg*(5.4) and final potential enrgy=0.

Also, in pure rotational motion, kinetic energy=1/2I2, where I is moment of inertia and is angular velocity.

Moment of inertia of rod about its end= ml2/3, where m is mass and l is length.

So,kinetic energy of rod=1/2*(ml2/3)2.

Initially, angular velocity is zero => initial kinetic energy=0. Also,finally, kinetic energy=1/2*(ml2/3)2, where is the final angular velocity.

Now, mechanical energy,i.e., the sum of kinetic and potential energies, is conserved as only conservative forces are involved during motion.

So,final mechanical energy=initial mechanical energy.

=> final potential energy+final kinetic energy=initial potential energy+initial kinetic energy

=>0+1/2*(ml2/3)2=mg*(5.4)+0

=>l2/6*2=g*5.4

=>2=6*g*5.4/l2. Here, l=10.8 m.

So,2=6*9.8*5.4/(10.8*10.8)=2.72222

=>=(2.72222)1/2=1.65 rad/sec