What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The diameter of the hockey puck is 0.076 m.

The mass of the puck is

m = 0.15 kg.

The diameter of the puck is 0.076 m, therefore its radius is

r = 0.076/2 = 0.038 m

The sliding speed is

v = 0.5 m/s

The angular velocity is

ω = 8.4 rad/s

The rotational moment of inertia of the puck is

I = (mr²) / 2

= 0.5 * (0.15 kg) * (0.038 m) ²

= 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.

The translational KE is

KE₁ = (1/2) * m*v²

= 0.5 * (0.15 kg) * (0.5 m/s) ²

= 0.0187 j

The rotational KE is

KE₂ = (1/2) * I*ω²

= 0.5 * (1.083 x 10⁻⁴ kg-m²) * (8.4 rad/s) ²

= 0.0038 J

The total KE is

KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J