Question

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?

Answer 1

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

ω = 0.6×8

ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

Answer 2

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.