Answer:
The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia = [tex]I[/tex]
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = [tex]5I[/tex] (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
therefore,
for the first case, the K.E. is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
and for the second case, the K.E. is given as
[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]
[tex]KE = \frac{1}{10}Iw^{2}[/tex]
this is one-tenth the kinetic energy before its spinning characteristics were changed.
This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.
Relation between Kinetic energy and Moment of Inertia:
Now, let's consider moment of inertia = I and angular speed = ω
It is asked that what would be change in Kinetic energy if
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
The kinetic energy of a spinning body is given as:
[tex]K.E.=\frac{1}{2} I. w^2[/tex]
On substituting the values, we will get:
[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]
Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.
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