A string wrapped around a hub of radius Rpulls with force ${\mathbf{F}}_{\mathbf{T}}$ on an object that rolls without slipping along horizontal rails on "wheels" of radius$\mathbf{r}\mathbf{<}\mathbf{R}$. Assume a massmand rotational inertia I.

(a) Prove that the ratio of${\mathbf{F}}_{\mathbf{T}}$to the object's acceleration is negative. (Note: This object can't roll without slipping unless there is friction.) You can do this by actually calculating the acceleration from the translational and rotational second la was of motion, but it is possible to answer this part without such a "real" calculation.

(b) Verify that${\mathbf{F}}_{\mathbf{T}}$times the speed at which the string moves in the direction of ${\mathbf{F}}_{\mathbf{T}}$ (i.e., the power delivered by${\mathbf{F}}_{\mathbf{T}}$) equals the rate at which the translational and rotational kinetic energies increase. That is.${\mathbf{F}}_{\mathbf{T}}$does all the work in this system, while the "internal" force does none. (c) Briefly discuss how parts (a) and (b) correspond to behaviors when an external electric field is applied to a semiconductor.

The radius of hub is R.

The force on the hub is $F_T$.

The expression for acceleration in translational motion is given by, ${\mathit{a}}_{\mathbf{T}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{T}}\mathbf{-}\mathbf{f}}{\mathbf{m}}$.

…(1)

The expression for acceleration in rotational motion is given by, .

$\mathit{a}\mathbf{,}\mathbf{=}\mathbf{-}\left(\frac{F_{T}R-fr}{I}\right)$ …(2)

The expression for total kinetic energy is given by, $\mathit{K}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\omega }}^{\mathbf{2}}\left(I+m{r}^2\right)$.

…(3)

The expression for speed of string is given by,$\mathit{V}\mathbf{=}\mathit{\omega }\left(R-r\right)$ .

(a)

Equate equation (1) and (2).

$$\begin{gathered} \frac{F_{T-f}}{m}=-r\left(\frac{F_{TR-fr}}{I}\right) \\ {F}_{T}I-fI=-rmR{F}_T+f{r}^{2}m \\ f=F_T\left(\frac{I+mRr}{I+m{r}^2}\right) \\ >F_T \end{gathered}$$

Since friction is bigger than applied force so, the direction of acceleration will be negative because its direction is opposite the direction of force.

Therefore, it is proved that ratio of force to acceleration is negative.

Differentiate equation (3) with respect to t.

$$\begin{gathered} \frac{dKE}{dt}=\frac{1}{2}\left(I+m{r}^2\right)\frac{d\omega {\omega }^2}{dt} \\ =\left(I+m{r}^2\right)\omega \frac{d\omega }{dt} \\ \frac{dKE}{dt}\left(I+m{r}^2\right)\omega \alpha \left(\because \frac{d\omega }{dt}=\alpha \right) \end{gathered}$$

…(4)

The acceleration is calculated as shown below.

$$\begin{gathered} \alpha =-\frac{F_T-f}{rm} \\ =-\frac{F_T\frac{\left(I+mR\right)}{\left(I+m{r}^2\right)}{F}_T}{rm} \\ =\frac{\left(R-r\right)}{\left(I+m{r}^2\right)}{F}_T \end{gathered}$$

Substitute $\frac{\left(R-r\right)}{\left(I+m{r}^2\right)}{F}_T\to \alpha$ in equation (4).

$$\begin{gathered} \frac{dKE}{dt}=\left(I+m{r}^2\right)\omega \left(\frac{\left(R-r\right)}{\left(I+m{r}^2\right)}\right) \\ =F_T\omega \left(R-r\right) \\ =F_{Tv} \end{gathered}$$

Therefore, it is proved that $\frac{dKE}{dt}=F_{Tv}$.

(c)

The material that has the property of both a conductor and a non-conductor is called the semi-conductor.

The semiconductor has the electrical conductivity more than a non-conductor but more than that of a conductor.

In case of the semiconductor, the net effective mass of electron is negative and they move in the direction opposite to the direction of external force.

By the definition of effective mass, the electric force does the net work on a particle moving through medium.

Therefore, if electric field is applied to a semiconductor then, the electric force does the net work on a particle moving through medium.