Respuesta :
Answer:
The moment of inertia of the wheel is 0.593 kg-m².
Explanation:
Given that,
Force = 82.0 N
Radius r = 0.150 m
Angular speed = 12.8 rev/s
Time = 3.88 s
We need to calculate the torque
Using formula of torque
[tex]\tau=F\times r[/tex]
[tex]\tau=82.0\times0.150[/tex]
[tex]\tau=12.3\ N-m[/tex]
Now, The angular acceleration
[tex]\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}[/tex]
[tex]\dfrac{d\omega}{dt}=20.73\ rad/s^2[/tex]
We need to calculate the moment of inertia
Using relation between torque and moment of inertia
[tex]\tau=I\times\dfrac{d\omega}{dt}[/tex]
[tex]I=\dfrac{I}{\dfrac{d\omega}{dt}}[/tex]
[tex]I=\dfrac{12.3}{20.73}[/tex]
[tex]I= 0.593\ kg-m^2[/tex]
Hence, The moment of inertia of the wheel is 0.593 kg-m².
