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Expert-verified Found in: Page 293 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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$\mathbf{2}\mathbf{.}\mathbf{0}\mathbit{K}\mathbit{g}$ box that slides on a frictionless surface inclined at angle ${\mathbit{\theta }}{\mathbf{=}}{\mathbf{}}{\mathbf{20}}{\mathbf{}}{\mathbf{°}}$with the horizontal. The box accelerates down the surface at$\mathbf{2}\mathbf{.}\mathbf{0}{\mathbf{m}}{\mathbf{s}}}^{\mathbf{2}}$ . What is the rotational inertia of the wheel about the axle? Rotational inertia about wheel is $0.054{\text{kgm}}^{2}$

See the step by step solution

## Step 1: 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$20°$

## Step 2: 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:

${\text{F}}_{net}=\text{M}×\text{a}\phantom{\rule{0ex}{0ex}}\tau =\text{I}\alpha =\text{T}×\text{r}\phantom{\rule{0ex}{0ex}}\alpha =\frac{\text{a}}{\text{r}}$

## Step 3: Draw the Free Body Diagram ## Step 4: 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$\theta =20°$

$⇒{\text{F}}_{net}={\text{m}}_{a}\text{gsin}20-{\text{T}}_{A}$

According to Newton’s law,

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

So

$\begin{array}{rcl}{\text{m}}_{a}\text{a}& =& {\text{m}}_{a}\text{gsin}20-{\text{T}}_{A}\\ & ⇒& 2×2=2×9.8×\text{sin}20-{\text{T}}_{A}\\ & ⇒& {\text{T}}_{A}=2.7\text{N}\end{array}$

Here torque is due to tension so

$\tau =\text{r}×\text{T}=\text{I}×\alpha \phantom{\rule{0ex}{0ex}}⇒\text{r}×{\text{T}}_{A}=\text{I}×\frac{\text{a}}{\text{r}}\phantom{\rule{0ex}{0ex}}⇒\text{I}=\frac{{\text{T}}_{A}×{\text{r}}^{2}}{\text{a}}\phantom{\rule{0ex}{0ex}}⇒\text{I}=\frac{2.7×{0.2}^{2}}{2}\phantom{\rule{0ex}{0ex}}⇒\text{I}=0.054{\text{kgm}}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ ### Want to see more solutions like these? 