Answer:
Explanation:
Let v be the linear velocity , ω be the angular velocity and I be the moment of inertia of the the puck.
Kinetic energy ( linear ) = 1/2 mv²
Rotational kinetic energy = 1/2 I ω²
I = 1/2 m r² ( m and r be the mass and radius of the puck )
Rotational kinetic energy = 1/2 x1/2 m r² ω²
= 1/4 m v² ( v = r ω )
Total energy
= Kinetic energy ( linear ) + Rotational kinetic energy
= 1/2 mv² + 1/4 m v²
= 3/4 mv²
rotational K E / Total K E = 1/4 m v² / 3/4 mv²
= 1 /3
So 1 /3 rd of total energy is rotational K E.
The fraction of the disk's total kinetic energy due to the rotation is 1.
The linear kinetic energy of the disk is calculated as follows;
K.E = ¹/₂mv²
where;
The initial linear speed of the disk = 0 (it was initially at rest)
The initial kinetic energy of the disk;
K.E[tex]_i[/tex] = 0
The final kinetic energy of the disk is rotational kinetic energy due to its angular speed;
[tex]K.E_{rot} = \frac{1}{2} I \omega ^2[/tex]
The total kinetic energy is calculated as follows;
[tex]K_{tot} = K.E_i + K.E_{rot}\\\\K.E_{tot} = 0 + K.E_{rot}\\\\K.E_{tot} = K.E_{rot}[/tex]
The fraction of the disk's total kinetic energy due to the rotation is calculated as follows;
[tex]\frac{K.E_{rot}}{K.E_{tot}} = \frac{K.E_{rot}}{K.E_{rot}} = 1[/tex]
Thus, the fraction of the disk's total kinetic energy due to the rotation is 1.
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