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madam [21]
3 years ago
8

The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob

tained 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
Physics
1 answer:
Naddik [55]3 years ago
8 0

(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

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