Question

A bicycle wheel with a radius of 34 cm is given an angular acceleration of 2.64 rad/s62

Answer 1

Answer: $0.041 kg m^{2}$

Explanation:

The bicycle wheel can be modeled as a hoop, which moment of inertia $I$ is:

$I=mr^{2}$ (1) Where $m$ is the mass and $r$ is the radius.

However, since we are given the applied force $F$ and angular acceleration $\alpha$, we can use the following equation:

$rF=I \alpha$ (2)

Where:

$F=0.32 N$

$\alpha=2.64 rad/s^{2}$

$r=34 cm \frac{1 m}{100 cm}=0.34 m$

Isolating $I$:

$I=\frac{rF}{\alpha}$ (3)

$I=\frac{(0.34 m)(0.32 N)}{2.64 rad/s^{2}}$ (4)

Finally:

$I=0.041 kg m^{2}$ This is the bicycle wheel moment of inertia