Question

A sharpening stone, also called whetstone, is used for sharpening ferrous tools. A round whetstone is mounted on the shaft of an electric motor and its moment of inertia is I = 4.0 kg.M2. When turned on, the whetstone starts from rest and motor exerts a constant torque of 20 N.M on it. What is the kinetic energy of the system after 8 seconds? Group of answer choices

Answer 1

Answer:

3.2 kJ

Explanation:

Since torque, τ = Iα where I = moment of inertia of round whetstone = 4.0 kgm² and α = angular acceleration of round whetstone.

So, α = τ/I

Given that τ = 20 Nm,

α = τ/I

α = 20 Nm/4.0 kgm²

α = 5 rad/s²

Using the equation for rotational motion, the angular velocity, ω = ω₀ + αt

where ω₀ = initial angular speed of round whetstone = 0 rad/s (since it starts from rest), ω = final angular speed of round whetstone, α = angular acceleration of round whetstone = 5 rad/s² and t = time of rotation = 8 s.

So, substituting the values of the variables into the equation, we have

ω = ω₀ + αt

ω = 0 rad/s + 5 rad/s² × 8s

ω = 0 rad/s + 40 rad/s

ω = 40 rad/s

So, its kinetic energy change ΔK = K₂ - K₁ = 1/2Iω² - 1/2Iω₀²

where K₁ = initial kinetic energy of round whetstone = 0 J (since the stone starts from rest) and K₂ = final kinetic energy of round whetstone after 8 s

Substituting the values of the variables into the equation, we have

K₂ - K₁ = 1/2Iω² - 1/2Iω₀²

K₂ - 0 J = 1/2 × 4.0 kgm² × (40 rad/s)² - 1/2 × 4.0 kgm² × (0 rad/s)²

K₂ = 2.0 kgm² × 1600 rad²/s² - 0 J

K₂ = 3200 kgm²rad²/s² - 0 J

K₂ = 3200 J

K₂ = 3.2 kJ