Question

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

by applying a force of 0.32 N on the edge of the wheel. What is the wheel's moment of inertia?

Answer 1

Answer:
`0.041 kg m^(2)`

Step-by-step 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 (1 m)/(100 cm)=0.34 m`

Isolating
`I`:

`I=(rF)/(\alpha)` (3)

`I=((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