Answer :
Answer:
The translational kinetic energy is 225 J
The rotational kinetic energy is 225 J
Explanation:
Given;
mass of the wheel, m = 2-kg
linear speed of the wheel, v = 15 m/s
Transnational kinetic energy is calculated as;
E = ¹/₂MV²
where;
M is mass of the moving object
V is the velocity of the object
E = ¹/₂ x 2 x (15)²
E = 225 J
Rotational kinetic energy is calculated as;
E = ¹/₂Iω²
where;
I is moment of inertia
ω is angular velocity
[tex]E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2[/tex]
E = ¹/₂ x 2 x (15)²
E = 225 J
Thus, the translational kinetic energy is equal to rotational kinetic energy