Answer:
I=0.489 kg.m²
Explanation:
Given that
Initial speed of wheel (ω₁)= 0 rad/s
θ = 8.2 rev in t= 12 s
We know that
1 rev = 2π rad
8.2 rev = 16.4 π rad
θ =16.4 π rad
Lets take angular acceleration of wheel is α rad/s²
We know that if angular acceleration is constant
[tex]\theta=\omega_1t+\dfrac{1}{2}\alpha t^2[/tex]
Now by putting the values
[tex]\theta=\omega_1t+\dfrac{1}{2}\alpha t^2[/tex]
[tex]16.4\times \pi=0\times 12+\dfrac{1}{2}\alpha \times 12^2[/tex]
α =0.715 rad/s²
The final speed of the wheel at 12 s
ω₂=ω₁ + α t
ω₂ = 0 + 0.715 x 12
ω₂ =8.58 rad/s
The final kinetic energy of the wheel
[tex]KE=\dfrac{1}{2}I\omega ^2[/tex]
[tex]36=\dfrac{1}{2}I\times 8.58^2[/tex]
I=0.489 kg.m²
This is moment of inertia of the wheel .
