Answer:
The translational kinetic energy is 225 J
The rotational kinetic energy is 225 J
Step-by-step 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
![E = (1)/(2) I \omega^2\\\\E = (1)/(2) *mr^2*((v)/(r))^2\\\\E = (1)/(2) *mr^2*(v^2)/(r^2) \\\\E = (1)/(2)mv^2](https://img.qammunity.org/2021/formulas/physics/high-school/sy4jqk5a18w10etpjx4dg2ptgl8888an05.png)
E = ¹/₂ x 2 x (15)²
E = 225 J
Thus, the translational kinetic energy is equal to rotational kinetic energy