Consider a solid sphere and a solid disk with the same radius and the same mass

Consider a solid sphere and a solid disc with the
Please show the work to numbers 3 and 4 so I can follow!

Consider a solid sphere and a solid disk with the same radius and mass. Explain why the solid disk has a greater moment of inertia than the solid sphere, even though it has the same total mass and the same radius. Calculate the moment of inertia and the kinetic energy of rotation for the following objects rotating around a central axis (in units of Joules): a solid sphere with a mass of 200 grams and a radius of 5.0 cm rotating at an angular speed of 2.5 rad/second. a thin spherical shell with a mass of 200 grams and a radius of 5.0 cm rotating at an angular velocity of 2.5 rad/s. a solid cylinder with a mass of 200 grams and a radius of 5.0 cm rotating at an angular velocity of 2.5 rad/s. a hoop with a mass of 200 grams and a radius of 5.0 cm rotating at an angular speed of 2.5 rad/s.

Answer

of them)

moment of inertia of the solid sphere, I1=2/5*m*r^2

moment of inertia of solid disk, I2=1/2*m*r^2

here, I2>I1

because it depends on the distribution of the masses around the axis,

3)

a) moment of inertia of the solid sphere, I=2/5*m*r^2

I=2/5*200*10^-3*(5*10^-2)^2

I=2*10^-4 kg.m^2

rotational kinetic energy KE=1/2*I*w^2

KE=1/2*2*10^-4*(2.5)^2

=6.25*10^-4 J b) moment of inertia of the spherical shell, I=2/3*m*r^2

I=2/3*200*10^-3*(5*10^-2)^2

I=3.3*10^-4 kg.m^2

rotational kinetic energy KE=1/2*I*w^2

KE=1/2*3.3*10^-4*(2.5)^2

=10.31*10^-4J

c) moment of inertia of the solid cylinder, I=1/2*m*r^2

I=1/2*200*10^-3*(5*10^-2)^2

I=2.5*10^-4 kg.m^2

rotational kinetic energy KE=1/2*I*w^2

KE=1/2*2.5*10^-4*(2.5)^2

=7.81*10^-4J

d) moment of inertia of the spherical shell, I=m*r^2

I=200*10^-3*(5*10^-2)^2

I=5*10^-4 kg.m^2

rotational kinetic energy KE=1/2*I*w^2

KE=1/2*5*10^-4*(2.5)^2

=15.62*10^-4J

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