Question

Calculate the rotational kinetic energy in the motorcycle wheel if it’s angular velocity is 110 rad/s. Assum m= 14.5 kg, R1= 0.29 m, and R2= 0.32 m. Moment of inertia for the wheel I= Unit = KE rot =. Unit=

Answer 1

Mass (m) = 14.5 kg

Angular velocity (ω) = 110 rad/s

Inner radius (R₁) = 0.29 m

Outer radius (R₂) = 0.32 m

Rotational kinetic energy is given by,

K.E = ½*Iω²

I – Moment of inertia

ω – Angular velocity in rad/s

We know ω, so just need to calculate the moment of inertia (I)

For an ring of mass m with an inner radius R₁ and outer radius R₂ the moment of inertia is given by,

I = ½*m (R₁² + R₂²)

Substituting given values,

I = ½ *(14.5) * (0.29+ 0.32)

I = 1.352 Kg m²

K.E = ½*Iω²

K.E = ½ * (1.352)*(110)²

Ans - K.E = 8179.6 J

The rotational kinetic energy is, K.E = 8179.6 J