91Ó°ÊÓ

Given data can be listed below,

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

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.

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

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

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

Substituting the value of Iand$\mathrm{Ó\neg }$ in the above equation.

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

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,

$K{E}_{rot}=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

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

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

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