91Ó°ÊÓ

1. The radius of wheel A, $r_A$is $0.10 \text{m}$.
2. The radius of wheel C,$r_C$is$0.25 \text{m}$.
3. The angular acceleration of wheel A,${\mathrm{Î\pm }}_A$is$1.6 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.
4. The angular speed $\mathrm{Ó\neg }$ is $100 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.

By using the formulas for tangential acceleration and angular speed, we can find thetime needed for wheel C to reach an angular speed of

1. The tangential acceleration $a_t=\mathrm{Î\pm }r$
2. The angular speed $\mathrm{Ó\neg }$is$\mathrm{Ó\neg }=\mathrm{Î\pm }t$
3. Hint: If the belt does not slip, the linear speed at the two rims must be equal.

Since the belt does not slip, a point on the rim of wheel C has the same tangential acceleration as a point on rim of wheel A. This means that

${\mathrm{Î\pm }}_A{r}_A={\mathrm{Î\pm }}_C{r}_C$

Where, ${\mathrm{Î\pm }}_A$ is the angular acceleration of wheel A, and ${\mathrm{Î\pm }}_C$is the angular acceleration of wheel C. Thus,

${\mathrm{Î\pm }}_C=\frac{r_A}{r_C}{\mathrm{Î\pm }}_A$

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{Î\pm }}_C=\left(\frac{0.10 \text{m}}{0.25 \text{m}}\right)×1.6\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right. \\ =0.64\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right. \end{gathered}$$

Angular speed of wheel C given by

${\mathrm{Ó\neg }}_C={\mathrm{Î\pm }}_{C}t$

The time t for it to reach an angular speed$\mathrm{Ó\neg }=100 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.=10.5\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$starting from rest is given by

$t=\frac{{\mathrm{Ó\neg }}_C}{{\mathrm{Î\pm }}_C}$

Substitute all the value in the above equation.

$$\begin{gathered} t=\frac{{\displaystyle 10.5\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{{\displaystyle 0.64 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}} \\ =16 \text{s} \end{gathered}$$

Hence the time is,$16 \text{s}$ .