Question

The rotational inertia of a disk about its axis is 0.70kgm^2. When a 2.0 kg weight is added to its rim, 0.40m from the axis, the rotational intertia becomes ?

Answer 1

0.70+2*(0.40)^2 ?

Answer 2

yes that right ,, the correct answer was ~1 kg.m^2

Answer 3

we use ,, Iz` =Iz"com" + ML^2
right ?

Answer 4

the M.I of disk+the M.I of the ring

Answer 5

A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter
that is given by (2/5)MR^2. A light string of length 3R is attached to the surface and used to
suspend the sphere from the ceiling. Its rotational inertia about the point of attachment at
the ceiling is ?

(2/5)MR^2 + M(3R)^2 = (47/5)MR^2
but the answer was (82/5)MR^2 !! what the mistake i had ??

Answer 6

u used parallel axis theorem... \[I=I _{cm}+Mh ^{2}\]
here h= the distance between center of mass and the new axis...
here h=4R

Answer 7

so I=2/5MR^2+M(4R)^2=82/5MR^2

Answer 8

Oooh !! Thanx a lot ,, I appreciate it !

Answer 9

:)

Answer 10

:)