A string is wrapped around a disk of mass 2.1 kg (its density is not necessarily uniform). Starting from rest, you pull the string with a constant force of 9 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance 0.11 m, your hand has moved a distance of 0.28 m (Figure 9.51).

(a) At this instant, what is the speed of the center of mass of the disk? (b) At this instant, how much rotational kinetic energy does the disk have relative to its center of mass? (c) At this instant, the angular speed of the disk is 75 rad/s. What is the moment of inertia of the disk?

Given data can be listed below,

• Mass of disk,$\mathrm{M}=2.1\mathrm{kg}$
• Force,$\mathrm{F}=9\hspace{0.33em}\mathrm{N}$
• The center of the disk move distance,$\mathrm{x}=0.11\hspace{0.33em}\mathrm{m}$
• Distance move by unwrapped,$\mathrm{d}=0.28\hspace{0.33em}\mathrm{m}$
• Speed of disk is$\mathrm{\omega }=0.75\hspace{0.33em}\mathrm{rad}/\mathrm{s}$

The total work done on the disk is given as,

$\mathrm{W}=\mathrm{F}(\mathrm{d}+\mathrm{x})$

Substituting 9 N for F , 0.28 m for , and 0.11m for x in the above equation.

$$\begin{gathered} \mathrm{W}=\left(9\mathrm{N}\right)\left(0.11\mathrm{m}+0.28\mathrm{m}\right) \\ =3.51\mathrm{J} \end{gathered}$$

Part a)

By using conservation of energy, the total work is equal to the sum of translational kinetic energy and rotational kinetic energy.

$$\begin{gathered} \mathrm{W}={\mathrm{KE}}_{\mathrm{trans}}+{\mathrm{KE}}_{\mathrm{rot}} \\ =\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{2}{\mathrm{l\omega }}^2 \end{gathered}$$

WhereIis the moment of inertia, and vis the speed of the disk.

$$\begin{gathered} \mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2 \\ \mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{R}} \end{gathered}$$

Substituting the value of Iand$\omega$ in the above equation.

$$\begin{gathered} \mathrm{W}=\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{2}\times \frac{1}{2}{\mathrm{MR}}^2\times \frac{{\mathrm{v}}^2}{{\mathrm{R}}^2} \\ =\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{4}{\mathrm{Mv}}^2 \\ =\frac{3}{4}{\mathrm{Mv}}^2 \end{gathered}$$

Substituting the 3.51 J for W, and 2.1 kg for M in the above equation

$$\begin{gathered} \left(3.51\mathrm{J}\right)=\frac{3}{4}\left(2.1\mathrm{kg}\right){\mathrm{v}}^2 \\ \mathrm{v}=1.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the disk at the center of mass is 1.5 m/s.

Part b)

The translational kinetic energy of the disk is given as,

${\mathrm{KE}}_{\mathrm{trans}}=\frac{1}{2}{\mathrm{Mv}}^2$

Substituting 1.5 m/s for v, and 2.1 kg for M in the above equation.

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{trans}}=\frac{1}{2}\left(2.1\mathrm{kg}\right){\left(1.5\mathrm{m}/\mathrm{s}\right)}^2 \\ =2.36\mathrm{J} \end{gathered}$$

Now, the rotational kinetic energy of the disk relative to the center of mass is,

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{rot}}=\mathrm{W}-{\mathrm{KE}}_{\mathrm{trans}} \\ =3.51\mathrm{J}-2.36\mathrm{J} \\ =1.15\mathrm{J} \end{gathered}$$

Thus,the rotational kinetic energy of the disk relative to the center of mass is 1.15 J.

Part c)

The rotational kinetic energy of the disk is given as,

$K{E}_{rot}=\frac{1}{2}I{\omega }^2$

Substituting 75 rad/s for $\omega$, and 1.15 J for $K{E}_{rot}$in the above equation.

$$\begin{gathered} \left(1.15J\right)=\frac{1}{2}I\left(75rad/s\right) \\ I=4.09\times {10}^{-4}kg.m^2 \end{gathered}$$

Thus, the moment of inertia of the disk is localid="1657859923221" $4.09\times {10}^{-4}\mathrm{kg}.{\mathrm{m}}^2$