Question

A string is wrapped around a uniform disk of mass$\mathbf{}\mathbf{M}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{kg}\mathbf{}$and radius$\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{11}\mathbf{\hspace{0.33em}}\mathbf{m}$ (Figure 9.63). Attached to the disk are four low-mass rods of radius$\mathbf{b}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{,}$, each with a small mass$\mathbf{}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{\hspace{0.33em}}\mathbf{kg}$at the end (Figure 9.63). The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force$\mathbf{}\mathbf{F}\mathbf{=}\mathbf{21}\mathbf{\hspace{0.33em}}\mathbf{N}$. At the instant that the center of the disk has moved a distance$\mathbf{}\mathbf{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{026}\mathbf{\hspace{0.33em}}\mathbf{m}$, an additional length$\mathbf{w}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{092}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{}$of string has unwound off the disk. (a) At this instant, what is the speed of the center of the apparatus? Explain your approach. (b) At this instant, what is the angular speed of the apparatus? Explain your approach.

Short answer

$$\begin{gathered} \mathrm{a})0.6244\hspace{0.33em}\mathrm{m}/\mathrm{s} \\ \mathrm{b})59.81\hspace{0.33em}\mathrm{rad}/\mathrm{s} \end{gathered}$$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the uniform disk is,$\mathrm{M}=1.2\hspace{0.33em}\mathrm{kg}$ .
• The radius of the uniform disk is,$\mathrm{R}=0.11\hspace{0.33em}\mathrm{m}$ .
• The radius of the four low mass rods is,$\mathrm{b}=0.14\hspace{0.33em}\mathrm{m}$ .
• The mass of the for low mass is, $\mathrm{m}=0.4\mathrm{kg}$.
• The force needed to pull the string is,$\mathrm{F}=21\hspace{0.33em}\mathrm{N}$ .
• The distance that the center of the disk has moved is,$\mathrm{d}=0.026\hspace{0.33em}\mathrm{m}$ .
• The additional length of the string is,$\mathrm{w}=0.092\hspace{0.33em}\mathrm{m}$ .

Significance of the work-energy theorem for the string

This theorem illustrates that the change in the kinetic energy of an object is equal to the net work done on that object.

The expression for the work-energy theorem is given as follows,

$\mathrm{W}=\mathrm{\Delta K}.\mathrm{E}........\left(1\right)$

The equation of the work done gives the speed of the apparatus.

The torque for the object can be determined by taking the product of the moment of inertia and angular acceleration. It can be expressed as follows,

The torque of the apparatus is expressed as-

$\mathrm{T}=\mathrm{{\rm I}}\times \mathrm{\alpha }........\left(2\right)$

Here, T is the torque, I is the moment of inertia and a is the angular acceleration of the apparatus.

Determination of the speed of the center of the apparatus

(a)

The expression for work done is given as follows,

$W=F\cdot d$

Here, F is the force applied to the body and d is the displacement of the object due to the applied force.

The expression for the change in kinetic energy is as follows,

$\mathrm{\Delta K}.\mathrm{E}.=\frac{1}{2}(\mathrm{M}+4\mathrm{m}){\mathrm{v}}^2$

Substitute all the values in equation (1).

$$\begin{gathered} \mathrm{F}.\mathrm{d}=\frac{1}{2}\left(\mathrm{M}+4\mathrm{m}\right){\mathrm{v}}^2 \\ \mathrm{v}=\sqrt{\frac{2\times \mathrm{F}.\mathrm{d}}{\mathrm{M}+4\mathrm{m}}} \end{gathered}$$

For, $\mathrm{F}=21\hspace{0.33em}\mathrm{N},\mathrm{d}=0.026\hspace{0.33em}\mathrm{m},\mathrm{M}=1.2\hspace{0.33em}\mathrm{kg}\mathrm{and}\mathrm{m}=0.4\hspace{0.33em}\mathrm{kg}.$

$$\begin{gathered} \mathrm{V}=\sqrt{\frac{2\times 21\mathrm{N}\times {\displaystyle \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}.0.026\mathrm{m}}{1.2\mathrm{kg}+4\times 0.4\mathrm{kg}}} \\ =0.6244\hspace{0.33em}\mathrm{m}/\mathrm{s}. \end{gathered}$$

Thus, the speed of the center of the apparatus is $0.6244\hspace{0.33em}\mathrm{m}/\mathrm{s}$.

Determination of the angular speed of the apparatus

(b)

The expression for torque cam also be expressed as follows,

$\mathrm{T}=\mathrm{F}\times \mathrm{R}$

Here,R is the radius of the disk.

The moment of inertia for the uniform disk is given as follows,

$\mathrm{I}=\left(\frac{1}{2}{\mathrm{MR}}^2+4{\mathrm{mb}}^2\right)$

Here,b is radius of the four low mass rods.

Substitute all the values in equation (2).

$\mathrm{F}\times \mathrm{R}=\left(\frac{1}{2}{\mathrm{MR}}^2+4{\mathrm{mb}}^2\right)\mathrm{\alpha }$

Substituting all the values in the above equation,

$$\begin{gathered} 21\mathrm{N}·0.11\mathrm{m}=\left(\frac{1}{2}·1.2\mathrm{kg}.{\left(0.11\mathrm{m}\right)}^2+4.0.4\mathrm{kg}.{\left(0.14\mathrm{m}\right)}^2\right)\mathrm{a} \\ \mathrm{a}=\frac{\left(21\mathrm{N}\times {\displaystyle \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}\right).0.11\mathrm{m}}{\left(\frac{1}{2}·1.2\mathrm{kg}.{\left(0.11\mathrm{m}\right)}^2+4.0.4\mathrm{kg}.{\left(0.14\mathrm{m}\right)}^2\right)} \\ =\frac{2.31\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}{\left(7.26\times {10}^{-3}\mathrm{kg}\times {\mathrm{m}}^2+0.03136\mathrm{kg}\times {\mathrm{m}}^2\right)} \\ =59.81\mathrm{red}/{\mathrm{s}}^2 \end{gathered}$$

The equation of the angular speed can be written as,

$\omega ={\omega }_0+\alpha t$

Here, ${\omega }_0$is the initial angular speed and$\omega$ is the final angular speed.

For,${\mathrm{\omega }}_0=0,\mathrm{\alpha }=59.81\mathrm{rad}/{\mathrm{s}}^2\mathrm{and}\mathrm{t}=1\hspace{0.33em}\mathrm{s}.$

$$\begin{gathered} \mathrm{\omega }=0+59.81\mathrm{rad}/{\mathrm{s}}^2\times 1\mathrm{s} \\ =59.81\hspace{0.33em}\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the angular speed of the apparatus is$59.81\mathrm{rad}/\mathrm{s}$.