Answer: [tex]0.041 kg m^{2}[/tex]
Explanation:
The bicycle wheel can be modeled as a hoop, which moment of inertia [tex]I[/tex] is:
[tex]I=mr^{2}[/tex] (1) Where [tex]m[/tex] is the mass and [tex]r[/tex] is the radius.
However, since we are given the applied force [tex]F[/tex] and angular acceleration [tex]\alpha[/tex], we can use the following equation:
[tex]rF=I \alpha[/tex] (2)
Where:
[tex]F=0.32 N[/tex]
[tex]\alpha=2.64 rad/s^{2}[/tex]
[tex]r=34 cm \frac{1 m}{100 cm}=0.34 m[/tex]
Isolating [tex]I[/tex]:
[tex]I=\frac{rF}{\alpha}[/tex] (3)
[tex]I=\frac{(0.34 m)(0.32 N)}{2.64 rad/s^{2}}[/tex] (4)
Finally:
[tex]I=0.041 kg m^{2}[/tex] This is the bicycle wheel moment of inertia
