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
[tex]\tau=I\alpha[/tex]
[tex]\alpha=\dfrac{\tau}{I}[/tex]
Where, I = moment of inertia
[tex]\alpha[/tex] = angular acceleration
[tex]\tau[/tex] = torque
Put the value into the formula
[tex]\alpha=\dfrac{3.00}{5.00}[/tex]
[tex]\alpha=0.6\ rad/s^2[/tex]
We need to calculate the final angular velocity
Initially wheel at rest so initial velocity is zero.
Using formula of angular velocity
[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]
[tex]\omega_{f}=\omega_{i}+\alpha t[/tex]
Put the value into the formula
[tex]\omega_{f}=0+0.6\times8.00[/tex]
[tex]\omega_{f}=4.8\ rad/s[/tex]
We need to calculate the rotational kinetic energy of the wheel
Using formula of the rotational kinetic energy
[tex]K.E_{rot}=\dfrac{1}{2}I\omega^2[/tex]
[tex]K.E_{rot}=\dfrac{1}{2}\times5.00\times(4.8)^2[/tex]
[tex]K.E_{rot}=57.6\ J[/tex]
Hence, The wheel's rotational kinetic energy is 57.6 J.
