16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

Given Information:

Torque = τ = 3.0 N. m

Moment of inertia = I = 5.0 kg. m²

Time = t = 8 seconds

Required Information:

Rotational kinetic energy = Erot = ?

Answer:

Rotational kinetic energy = 57.6 Joules

Explanation:

We know that the rotational kinetic energy of the wheel is the energy due to its rotation and is part of its total kinetic energy and is given by

Erot = ½Iω²

Where ω is the angular velocity and I is the moment of inertia of the wheel.

We also know the relation between torque and moment of inertia is

τ = Iα

Where α is the angular acceleration of the wheel.

α = τ/I

From the equations of kinematics, we know that final angular velocity is given by

ω = ω₀ + αt

Where ω₀ is the initial angular velocity of the wheel and since wheel starts from rest, ω₀ is zero.

ω = 0 + αt

ω = αt

ω = (τ/I) t

ω = τ*t/I

Finally the equation of rotational kinetic energy becomes

Erot = ½Iω²

Erot = ½I (τ*t/I) ²

Erot = ½*5 * ((3*8) / 5) ²

Erot = ½*5 * (23.04)

Erot = ½ * (115.2)

Erot = 57.6 J

Therefore, the wheel's rotational kinetic energy at the end of 8 s is 57.6 Joules.