Question

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 73.0 N is applied to the rim of the wheel. The wheel has radius 0.110 m. Starting from rest, the wheel has an angular speed of 14.9 rev/s after 3.41 s. What is the moment of inertia of the wheel?

Answer 1

Approximately
`0.293\; {\rm kg \cdot m^(2)}`.

Step-by-step explanation:

Since the wheel started from rest, initial angular velocity will be
`\omega_(0) = 0\; {\rm rad \cdot s^(-1)}`. It is given that the angular velocity
`\omega_(1)` is
`14.9\; {\rm rev \cdot s^(-1)}` after
`t = 3.41\; {\rm s}`. Apply unit conversion and ensure that all angular velocity are measured in radians-per-second:

`\begin{aligned} \omega_(1) &= 14.9\; {\rm rev \cdot s^(-1)} * \frac{2\, \pi \; {\rm rad}}{1\; {\rm rev}} \\ &\approx 93.620\; {\rm rad \cdot s^(-1)}\end{aligned}`.

Change in angular velocity:

`\begin{aligned} \Delta \omega = \omega_(1) - \omega_(0) \approx 93.620\; {\rm rad \cdot s^(-1)}\end{aligned}`.

Since the tangential force is constant and there is no friction on the wheel, the angular acceleration
`\alpha` of this wheel will be constant. Since the change in velocity
`\Delta \omega \approx 93.620\; {\rm rad \cdot s^(-1)}` was achieved within
`t = 3.41\; {\rm s}`, the average angular acceleration will be:

`\begin{aligned} \alpha &= (\Delta \omega)/(t) \\ &\approx \frac{93.620\; {\rm rad \cdot s^(-1)}}{3.41\; {\rm s}} \\ &\approx 27.45\; {\rm rad \cdot s^(-2)}\end{aligned}`.

At a distance of
`r = 0.110\; {\rm m}` from the axis of rotation, the tangential force
`F = 73.0\; {\rm N}` will exert on the wheel a torque
`\tau` of magnitude:

`\begin{aligned} \tau &= F\, r \\ &= (73.0\; {\rm N})\, (0.110\; {\rm m}) \\ &\approx 8.030\; {\rm N \cdot m}\end{aligned}`.

Let
`I` denote the moment of inertia of this wheel. The equation
`\alpha = (\tau / I)` relates angular acceleration
`\alpha` to moment of inertia
`I\!` and net torque
`\tau`. Rearrange this equation to find the moment of inertia:

`\begin{aligned}I &= (\tau)/(\alpha) \\ &\approx \frac{8.030\; {\rm N\cdot m}}{27.45\; {\rm rad \cdot s^(-2)}} \\ &\approx 0.293\; {\rm N \cdot m \cdot s^(2)} \\ &= 0.293 \; {\rm kg \cdot m^(2)}\end{aligned}`.

Note that the unit "radians" is typically ignored. Additionally,
`1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^(-2)}`.

Hence, the moment of inertia of this wheel is approximately
`0.293\; {\rm kg \cdot m^(2)}`.