Question

In Fig.$\mathbf{10}\mathbf{}\mathbf{-}\mathbf{}\mathbf{55}$, a wheel of radius $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{}\mathbf{\text{m}}\mathbf{}$is mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a$\mathbf{2}\mathbf{.}\mathbf{0}\mathit{K}\mathit{g}$box that slides on a frictionless surface inclined at angle $\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{20}\mathbf{}\mathbf{°}$with the horizontal. The box accelerates down the surface at$\mathbf{2}\mathbf{.}\mathbf{0}{\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.}^{\mathbf{2}}$. What is the rotational inertia of the wheel about the axle?

Short answer

Rotational inertia about wheel is$0.054{\text{kgm}}^2$

Step-by-step solution

Given

1. Radius of wheel is$0.2\text{m}$
2. Mass of box is$2.0\text{kg}$
3. Acceleration is$2.0\text{m}/{\text{s}}^2$
4. Angle of inclination is$20°$

To understand the concept

Use Newton’s second law to find tension. Using the formula for the torque in terms of tension and radius and also using the formula in terms of moment of inertia and angular acceleration, find the acceleration.

Formula:

$$\begin{gathered} {\text{F}}_{net}=\text{M}\times \text{a} \\ \tau =\text{I}\alpha =\text{T}\times \text{r} \\ \alpha =\frac{\text{a}}{\text{r}} \end{gathered}$$

Draw the Free Body Diagram

Calculate the rotational inertia of the wheel about the axle

Net force acting on block is as follow

${\text{F}}_{net}={\text{m}}_a\text{gsin}\theta -{\text{T}}_A$

Here$\theta =20°$

$\Rightarrow {\text{F}}_{net}={\text{m}}_a\text{gsin}20-{\text{T}}_A$

According to Newton’s law,

${\text{F}}_{net}={\text{m}}_a\text{a}$

So

$\begin{array}{rcl}{\text{m}}_a\text{a}& =& {\text{m}}_a\text{gsin}20-{\text{T}}_A\\ & \Rightarrow & 2\times 2=2\times 9.8\times \text{sin}20-{\text{T}}_A\\ & \Rightarrow & {\text{T}}_A=2.7\text{N}\end{array}$

Here torque is due to tension so

$$\begin{gathered} \tau =\text{r}\times \text{T}=\text{I}\times \alpha \\ \Rightarrow \text{r}\times {\text{T}}_A=\text{I}\times \frac{\text{a}}{\text{r}} \\ \Rightarrow \text{I}=\frac{{\text{T}}_A\times {\text{r}}^2}{\text{a}} \\ \Rightarrow \text{I}=\frac{2.7\times {0.2}^2}{2} \\ \Rightarrow \text{I}=0.054{\text{kgm}}^2 \end{gathered}$$