Relation Between Inertia And Kinetic Energy Question

[zinedine_88]

OK...


Please consider that picture in the following link

http://images.google.com/imgres?img...ages?q=moment+of+inertia&gbv=2&svnum=10&hl=en


It says that the DISK will reach the ground more quickly than the HOOP, since it has smaller moment of Inertia.! I GET THAT!

But what I cannot understand is, if that is true, how come when you calculate the kinetic energies of the DISK and the HOOP according to the formula KE=1/2mv^2,
that the kinetic energy of the disk is twice less than the kinetic energy of the HOOP.

put in a more simple words, HOW COME THE OBJECT THAT REACHES THE GROUND FIRST, HAS LESS KINETIC ENERGY than the slower object?

PLEASE EXPLAIN me!

Tony

 

[jtbell]

Both objects actually have the same total kinetic energy, but they divide it differently between translational kinetic energy ($\frac{1}{2}mv^2$) and rotational kinetic energy. Look up rotational kinetic energy in your textbook and see how it is related to moment of inertia.

 

[Doc Al]

But what I cannot understand is, if that is true, how come when you calculate the kinetic energies of the DISK and the HOOP according to the formula KE=1/2mv^2,
that the kinetic energy of the disk is twice less than the kinetic energy of the HOOP.

As jtbell stated, 1/2mv^2 is the formula for translational KE. You are probably talking about rotational KE. For a given translational speed, a hoop has twice the rotational KE as does a cylinder (if the mass is the same).

put in a more simple words, HOW COME THE OBJECT THAT REACHES THE GROUND FIRST, HAS LESS KINETIC ENERGY than the slower object?

Whichever object reaches the ground first must have greater speed and thus more translational KE (if the mass is the same). Since the total KE is the same for both (again as jtbell explained), and since the cylinder has less rotational KE it will have more translational KE and thus will reach the ground sooner.