Question

Figure $\mathbf{10}\mathbf{-}\mathbf{43}\mathbf{}$ shows a uniform disk that can rotate around its center like a merry-go-round. The disk has a radius of $\mathbf{2.00}\mathbf{\text{\hspace{0.17em}cm}}\mathbf{}\mathbf{}$ and a mass of $\mathbf{20.0}\mathbf{\text{\hspace{0.17em}grams}}\mathbf{}\mathbf{}$and is initially at rest. Starting at time $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}$, two forces are to be applied tangentially to the rim as indicated, so that at time $\mathit{t}\mathbf{=}\mathbf{}\mathbf{1.25}\mathbf{\text{\hspace{0.17em}s}}$the disk has an angular velocity of$\mathbf{250}\mathbf{}\mathbf{\text{rad}}\mathbf{/}\mathbf{\text{s}}\mathbf{}$ counterclockwise. Force ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$has a magnitude of$\mathbf{0}\mathbf{.100}\mathbf{\text{\hspace{0.17em}N}}$ . What is magnitude${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$ ?

Short answer

The magnitude of force, ${\text{F}}_{\text{2}}$is, $0140\text{\hspace{0.17em}N}$.

Step-by-step solution

Understanding the given information

1. The mass of disk is,$\mathrm{M}=0.02\text{\hspace{0.17em}kg}$
2. The mass of disk is,$\mathrm{R}=0.02\text{\hspace{0.17em}m}$
3. The angular velocity is,$\mathrm{\omega }=250\text{\hspace{0.17em}rad}/\text{s}$
4. The time is,$t=1.25\text{\hspace{0.17em}s}$
5. The force is, $F_1=0.100\text{\hspace{0.17em}N}$

Concept and formula used in the given question

There are two forces acting on the disk, you can write the equation for net torque caused by these two forces. Using the given data, you can find the angular acceleration of the disk. Using this angular acceleration, you can find second force. The formulas used are given below.

Net Torque ${\mathit{\tau }}_{\mathbf{N}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{I}\mathit{\alpha }$

role="math" localid="1660911730037" $\mathit{R}{\mathit{F}}_{\mathbf{2}}\mathbf{-}\mathit{R}{\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathit{I}\mathit{\alpha }$

Where,
$\mathbf{\alpha }\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{t}}$

Moment of inertia
$\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{MR}}^{\mathbf{2}}$

Calculation for the magnitude of  F→2

We have,

${\mathrm{RF}}_1=\mathrm{I\alpha }$ …(1)

Where,
$\mathrm{\alpha }=\frac{\mathrm{\omega }}{\mathrm{t}}$

Also, moment of inertia can be given by formula,

$\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2$

Putting both the values in equation (1), we get,

$\begin{array}{c}{\mathrm{RF}}_2-{\mathrm{RF}}_1=\frac{1}{2}{\mathrm{MR}}^2\times \frac{\mathrm{\omega }}{\mathrm{t}}\\ \mathrm{R}({\mathrm{F}}_2-{\mathrm{F}}_1)=\frac{1}{2}{\mathrm{MR}}^2\times \frac{\mathrm{\omega }}{\mathrm{t}}\\ 2({\mathrm{F}}_2-{\mathrm{F}}_1)=\frac{\mathrm{MR\omega }}{\mathrm{t}}\end{array}$

i.e,

${\mathrm{F}}_2=\frac{\mathrm{MR\omega }}{2\mathrm{t}}+{\mathrm{F}}_1$

Putting values in above equation, we get,

$\begin{array}{c}{\mathrm{F}}_2=\frac{\left(0.02\text{\hspace{0.17em}kg}\right)\left(0.02\text{\hspace{0.17em}m}\right)(250\text{\hspace{0.17em}rad}/\text{s})}{2\left(1.25\text{\hspace{0.17em}s}\right)}+\left(0.100\text{\hspace{0.17em}N}\right)\\ =0.140\text{\hspace{0.17em}N}\end{array}$

Hence the magnitude of $F_2$ is, $0.140\text{\hspace{0.17em}N}$.