Question

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:

The wheel's rotational kinetic energy is 57.6 J.

Explanation:

Given that,

Moment of inertia = 5.00 kg.m²

Torque = 3.00 N.m

Time = 8.00 s

We need to calculate the angular acceleration

Using formula of the torque act on the wheel

$\tau=I\alpha$

$\alpha=\dfrac{\tau}{I}$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\tau$ = torque

Put the value into the formula

$\alpha=\dfrac{3.00}{5.00}$

$\alpha=0.6\ rad/s^2$

We need to calculate the final angular velocity

Initially wheel at rest so initial velocity  is zero.

Using formula of angular velocity

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}$

$\omega_{f}=\omega_{i}+\alpha t$

Put the value into the formula

$\omega_{f}=0+0.6\times8.00$

$\omega_{f}=4.8\ rad/s$

We need to calculate the rotational kinetic energy of the wheel

Using formula of the rotational kinetic energy

$K.E_{rot}=\dfrac{1}{2}I\omega^2$

$K.E_{rot}=\dfrac{1}{2}\times5.00\times(4.8)^2$

$K.E_{rot}=57.6\ J$

Hence, The wheel's rotational kinetic energy is 57.6 J.