question archive Consider a solid sphere and a solid disk with the same radius and the same mass
Consider a solid sphere and a solid disk with the same radius and the same mass
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Consider a solid sphere and a solid disk with the same radius and the same mass. Explain why the solid disk has a greater moment of inertia than the solid sphere, even though it has the same overall mass and radius. Calculate the moment of inertia and the rotational kinetic energy for the following objects spinning about a central axis (in units of Joules): a solid sphere with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a thin spherical shell with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a solid cylinder with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a hoop with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec.
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2)
moment of inertia of 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 depend the distribution of mass around the axis,
3)
a)
moment of inertia of 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 kinteic energy K.E=1/2*I*w^2
K.E=1/2*2*10^-4*(2.5)^2
=6.25*10^-4 J
b)
moment of inertia of 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 kinteic energy K.E=1/2*I*w^2
K.E=1/2*3.3*10^-4*(2.5)^2
=10.31*10^-4 J
c)
moment of inertia of 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 kinteic energy K.E=1/2*I*w^2
K.E=1/2*2.5*10^-4*(2.5)^2
=7.81*10^-4 J
d)
moment of inertia of spherical shell, I=m*r^2
I=200*10^-3*(5*10^-2)^2
I=5*10^-4 kg.m^2
rotational kinteic energy K.E=1/2*I*w^2
K.E=1/2*5*10^-4*(2.5)^2
=15.62*10^-4 J
