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Q87P

Expert-verifiedFound in: Page 293

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$\mathbf{2}\mathbf{.}\mathbf{0}\mathit{K}\mathit{g}$ box that slides on a frictionless surface inclined at angle ${\mathit{\theta}}{\mathbf{=}}{\mathbf{}}{\mathbf{20}}{\mathbf{}}{\mathbf{\xb0}}$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?**

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

- Radius of wheel is$0.2\text{m}$
- Mass of box is$2.0\text{kg}$
- Acceleration is$2.0\text{m}/{\text{s}}^{2}$
- Angle of inclination is$20\xb0$

**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:**

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

Net force acting on block is as follow

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

Here$\theta =20\xb0$

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

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

$\tau =\text{r}\times \text{T}=\text{I}\times \alpha \phantom{\rule{0ex}{0ex}}\Rightarrow \text{r}\times {\text{T}}_{A}=\text{I}\times \frac{\text{a}}{\text{r}}\phantom{\rule{0ex}{0ex}}\Rightarrow \text{I}=\frac{{\text{T}}_{A}\times {\text{r}}^{2}}{\text{a}}\phantom{\rule{0ex}{0ex}}\Rightarrow \text{I}=\frac{2.7\times {0.2}^{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \text{I}=0.054{\text{kgm}}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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