Question

The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is obtained by using integral calculus to add up the contributions of all the atoms in the disk. The factor of 1/2 reflects the fact that some of the atoms are near the center and some are far from the center; the factor of 1/2 is an average of the square distances. A uniform-density disk whose mass is 16 kg and radius is 0.19 m makes one complete rotation every 0.2 s. (a) What is the moment of inertia of this disk? Entry field with correct answer 0.2888 kg · m2 (b) What is its rotational kinetic energy? Entry field with correct answer 142.52 J (c) What is the magnitude of its rotational angular momentum? Entry field with incorrect answer now contains modified data kg · m2/s

Answer 1

(a) $0.2888 kg m^2$

The moment of inertia of a uniform-density disk is given by

$I=\frac{1}{2}MR^2$

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

$I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2$

(b) 142.5 J

The rotational kinetic energy of the disk is given by

$K=\frac{1}{2}I\omega^2$

where

I is the moment of inertia

$\omega$ is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s$

And so, the rotational kinetic energy is

$K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J$

(c) $9.07 kg m^2 /s$

The rotational angular momentum of the disk is given by

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular velocity

Substituting the values found in the previous parts of the problem, we find

$L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s$