Studying for an exam, need an explanation for inertia and rotational kinematics?

by Doc_Smith on December 31, 2009

{ 1 comment… read it below or add one }

Edward December 31, 2009 at 8:51 pm

I don’t know…let’s see

Ke= 0.5Iw^2

I1= I2
w1=w2

so why would it it be different? Unless…

the moment of inertia for a sphere
I (sphere)=Is= (2/5) m R^2
and
I(disk)= Id= (1/2) m R^2

The total kinetic energy = Ke( traslational) +Ke(rotational)

Ke( translational) =(1/2)mV^2
since V=wR and
m(sphere)= (5/2) I /R^2
m(disk)= 2 I /R^2

Ke(translational) =(1/2)m (wR)^2

Ke(translational sphere) =[(1/2)(5/2) I /R^2](wR)^2=
Ke(translational sphere) =(5/4) I /(w)^2=

Ke(translational disk) =[(1/2)(2) I /R^2]wR)^2
Ke(translational disk) = I /(w)^2

Ke(translational sphere)/Ker(translational disk)= =(5/4) I /(w)^2 / I /(w)^2= 5/4

The linear kinetic energy will be larger for a sphere.

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